• What to choose for u: L.I.A.T.E. (pick the one higher on the list)
  1. Logarithms (e.g., \ln x)
  2. Inverse trig (e.g., arctan(x))
  3. Algebraic (e.g., x^2)
  4. Trig (e.g., \sin(x))
  5. Exponential (e.g., e^x)

Deriving the Integration by Parts Formula:

\text{Product Rule: } \frac{d}{dx} [ f(x)g(x) ] = f(x)g'(x) + f'(x)g(x)

Now, let u = f(x) and v = g(x).

\text{Solve: } \int x \sin x dx

Let u = x and dv = \sin x dx.

• (Now we want to find v and du to use Integration by Parts.)

1. To find v we can take the integral of both sides of the dv equation:

dv = \sin x dx \\ \int dv = \int \sin x dx \\ v = -\cos x

2. To find du we can take the derivative of both sides of the u equation:

u = x \\ u' = x' \\ du = 1

3. Now we can plug into Integration by Parts:

\begin{aligned} \int u dv &= uv - \int v du \\ \int x \sin x dx &= x ( - \cos x ) - \int - \cos x dx \end{aligned}

4. Now, solve the last integral on the right to get the final answer:

\int x \sin x dx = - x \cos x + \sin x + c

Tip: You can verify the answer seeing if (- x \cos x + \sin x)' = x \sin x

\text{Solve: } \int (\cos x)^3 dx

Notes:

• U-sub doesn’t simplify this because there isn’t a \sin x to cancel out the \frac{du}{dx}
• Integration by parts make the problem more difficult in every step.
• Because of this, we must resort to trigonometric substitution.

Recall the following identity: \begin{aligned} \boxed{ \text{Pythagorean Identity: } \cos^2 x + \sin^2 x = 1 } \end{aligned}

It logically follows that: \begin{aligned} \cos^2 x &= 1 - \sin^2 x \\ \sin^2 x &= 1 - \cos^2 x \end{aligned}

Note: For this problem, we only need to use the \cos^2 x identity, but I’m listing both for demonstration.

\begin{aligned} \int \sin^2 x dx &= \int ( \frac{1}{2} - \frac{1}{2} \cos 2x ) dx \\ &= \frac{1}{2} - \frac{1}{2} \cos 2x dx \\ &= \frac{1}{2} x - \frac{1}{4} \sin 2x + C \end{aligned}

Note: Half-angle identities are also-known-as power-reduction formulas.

• If you don’t know them, you might end up stuck!

\text{Solve: } \int \sqrt{9-x^2} dx

Solving:

\text{Identity: } a^2 \cos^2 \theta = a^2 - a^2 \sin^2 \theta \\

\text{Let } x = 3 \sin \theta \\ \text{Then } dx = 3 \cos \theta d \theta

Now we can substitute our x and dx: \int \sqrt{9 - x^2} dx = \int \sqrt{ 9 - (3 \sin \theta )^2 } 3 \cos \theta d \theta \\ = \int \sqrt{9 - 9 \sin^2 \theta} 3 \cos \theta d \theta \\~\\ \text{Apply Pythagorean identity: } \\~\\ = \int \sqrt{9 \cos^2 \theta} 3 \cos \theta d \theta \\ = \int 3 \cos \theta \times 3 \cos \theta d \theta \\ = 9 \int \cos^2 \theta d \theta \\~\\ \text{Half-angle identity: } \\~\\ = 9 \int \frac{1 + \cos (2 \theta) }{2} d \theta \\ = 9 ( \frac{1}{2} \theta + \frac{\sin (2 \theta) }{4} \theta ) + C\\ = \frac{9}{2} \theta + \frac{9}{4} \sin (2 \theta) +C \\~\\ \text{Map $\theta$ to $x$ with inverse trig: } \\~\\ x = 3 \sin\theta \to \frac{x}{3} = \sin \theta \to \text{Inverse sine} \to \arcsin \frac{x}{3} = \theta \\~\\ \text{We need to manipulate $\sin 2 \theta$ to use the mapping: } \\~\\ \boxed{ \text{Double-Angle Identitities: } \begin{aligned} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ \cos 2 \theta &= \cos^2 \theta - \sin^2 \theta \end{aligned} } \\~\\ \text{Applying the double-angle identity: } \\~\\ \frac{9}{2} \theta + \frac{9}{4} \sin (2 \theta) +C = \frac{9}{2} \theta + \frac{9}{4} ( 2 \sin \theta \cos \theta ) \\~\\ \text{Now we need just to find $\cos \theta$: } \\~\\ \text{Since $\sin \theta = \frac{x}{3}$}:

Tip: The square root you get from solving the missing side of the triangle should match the square root in the original problem.

• If it doesn’t match, you did something wrong.

\text{Solve: } \int \frac{3x+2}{x^3 - x^2 - 2x} dx

1. Factor the denominator into distinct linear factors.

= \int \frac{3x+2}{ x ( x^2 - x^1 - 2 )} dx = \int \frac{3x+2}{ x ( x-2) ( x+1 )} dx

2. Split the fraction into multiple tiny fractions.
   • (Keep numerators as variables for now)

= \int \frac{3x+2}{ x ( x-2) ( x+1 )} dx = \int \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} dx

3. Multiply both sides of the equation by the denominator.

\int 3x+2 dx = \int A(x-2)(x+1) + B(x)(x+1) + C(x)(x-2) dx

4. Get coefficients with strategic substitution
   • (We do this by using various values of x (roots of denominators) to get equations for each coefficient.)

$$

5. Plug coefficients back in.
   • (Use the form from step 2)

= \int \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+1} dx = \int \frac{-1}{x} + \frac{\frac{4}{3}}{x-2} - \frac{\frac{1}{3}}{x+1} dx

6. Solve.
   • (See “Integrals for Partial Fractions” below)

\text{Solve: } \frac{x^2 + 3x + 5}{x + 1} dx

We can simplify this by using long division.

x^2 + 3x + 5 \div x + 1 = x + 2 \text{r} 3 \\~\\ \text{Therefore: } \frac{x^2 + 3x + 5}{x + 1} dx = \int x + 2 + \frac{3}{x+1} dx

Thus, the answer is: \int x + 2 + \frac{3}{x+1} dx = \frac{x^2}{2} + 2x + 3 \ln |x + 1| + C

Tip: Remember, when performing long division the result is simply the quotient + remainder / divisor.

\text{Solve: } \int \frac{x-2}{(2x-1)^2 (x-1)} dx

You need to split this in a special way to handle the exponent: \int \frac{x-2}{(2x-1)^2 (x-1)} dx = \int \frac{A}{2x-1} + \frac{B}{(2x-1)^2} + \frac{C}{x-1} dx

Now we multiply both sides by the denominator to get the following: \int x-2 dx = \int A(2x-1)(x-1) + B(x-1) + C(2x-1)^2 dx

$$

Plug back in coefficients: \int \frac{2}{2x-1} + \frac{3}{(2x-1)^2} - \frac{1}{x-1} dx

• If given a problem to solve a y-function from y=a to y=b, simply solve for x and x', and plug it into the above formula, replacing f(x) with x, f'(x) with x', and dx with dy.

\text{Because } x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

• M: Moment of system
• m: Total mass

If objects are on a 2D plane, we get the center of mass by calculating the center mass in the x and y-directions separately.

• M_y: Moment ith respect to y-axis
• M_x: Moment with respect to x-axis
• m: Total mass

\begin{aligned} m_1 &= 2 \text{kg}, (-1,3) \\ m_2 &= 6 \text{kg}, (1,1) \\ m_3 &= 4 \text{kg}, (2,-2) \\ \end{aligned} \\~\\ \begin{aligned} x &= \frac{ 2(-1) + 6(1) + 4(2) }{ 2 + 6 + 4 } \\ &= \frac{-2 + 6 + 8}{12} \\ &= 1 \end{aligned} \\~\\ \begin{aligned} y &= \frac{ 2(3) + 6(1) + 2(-2) }{ 2+6+4 } \\ &= \frac{6+6-8}{12} \\ &= \frac{1}{3} \end{aligned}

• Therefore, the center of mass is (1, \frac{1}{3})

Now, what if mass is evenly spread continuously throughout a 2D sheet of metal, called a lamina?

• (We’ll assume the density is constant throughout, but the sheet can be a nonstandard shape)

Consider the lamina whose shape is a region bounded by y=f(x) above, x-axis below, and lines x=a and x=b.

Example: Deriving center of mass formula for a rectangle

• The center of mass is right the middle.
• Symmetry Principle: If region R is symmetric about a straight line, its geometric center lies on l.

1. First, let’s get the center of mass (x_i^* is the center of the rectangle) \text{Center of Mass of Rectangle: $(x_i^*, \frac{f(x_i^*}{2})$} \\~\\ \text{x-value: } x_i^* = \frac{x_i + x_i + 1}{2} \qquad \text{y-value: } \frac{f(x_i^*}{2}

2. Now, let’s find the density (\rho).

• (Recall that mass is density \times area.) \text{Mass of Rectangle: } m = \int_a^b \rho f(x) dx

3. Now we need moment with respect to the x-axis and y-axis. \text{Moments for One Rectangle: } \\~\\ \text{I. W.r.t. y-axis: } \rho f(x_i^*) \Delta x \times x_i^* \\~\\ \text{II. W.r.t. x-axis: } \rho \frac{[f(x_i^*)]^2}{2} \Delta x

4. So, M_y = \sum \rho f(x_i^*) \Delta x \times x_i^* \\ \\~\\ M_x = \sum \rho \frac{[f(x_i^*)]^2}{2} \Delta x

5. Which can be rewritten as: M_y = \int_a^b \rho x f(x) dx \\~\\ M_x = \int_a^b \rho \frac{[f(x_i^*)]^2}{2} dx

Q: Find center of mass of a lamina bounded by f(x)=\sqrt{x}, x-axis, over [0,4]

A:

1. Let’s start by finding the total mass (m) \begin{aligned} m &= \int_0^4 \rho \sqrt{x} dx \\ &= \rho \int_0^4 \sqrt{x} dx \\ &= \rho \frac{2}{3} ( 4^{3/2} - 0 ) \\ &= \frac{2 \rho}{3} (2)^3 \\ m &= \frac{16 \rho}{3} \end{aligned}

2. Next let’s find the moment with respect to the y-axis

\begin{aligned} M_y &= \int_0^4 \rho x \sqrt{x} dx \\ &= \rho \int_0^4 x^{3/2} dx \\ &= \rho \times \frac{2}{5} \times (\sqrt{x})^5 \\ M_y &= \frac{64 \rho}{5} \end{aligned}

3. And now, M_x:

\begin{aligned} M_x &= \int_0^4 \rho \frac{ \sqrt{x}^2 }{ 2 } dx \\ &= \frac{\rho}{2} \int_0^4 x^2 dx \\ &= \frac{\rho}{4} (4^0 - 0^2) \\ M_x &= 4 \rho \end{aligned}

4. Finally:

\begin{aligned} x &= \frac{\frac{64 \rho}{5}}{\frac{16 \rho}{3}} \\ &= \frac{12}{5} \end{aligned} \\~\\ \begin{aligned} y &= \frac{4 \rho}{\frac{16 \rho}{3}} \\ &= \frac{3}{4} \end{aligned}

5. Therefore, the center of mass is (2.4, 0.75)

• Q: Find the center of mass of the region bounded between f(x)=1-x^2 and g(x)=x-1

• A:

1. We must find a and b, so we’ll set the equations equal to each other and solve the roots: \begin{aligned} f(x) &= g(x) \\ 1 - x^2 &= x-1 \\ 0 &= x^2 + x -2 \\ 0 &= (x-1)(x+2) \\ \end{aligned}
   • Thus, the a and b is [-2, 1]
2. Now we can solve for M_y, M_x, and m.

\begin{aligned} m &= \int_{-2}^1 1-x^2 - (x-1) dx \\ &= \int_{-2}^1 1-x^2 - x + 1 dx \\ &= \int_{-2}^1 -x^2 -x + 2 dx \\ &= -\frac{x^3}{3} - \frac{x^2}{2} + 2x |_{-2}^1 \\ &= \frac{-9}{3} + \frac{3}{2} + 6 \\ &= 3 + \frac{3}{2} \\ &= \frac{9}{2} \end{aligned}

\begin{aligned} M_y &= \int_{-2}^1 x ( -x^2 -x + 2 ) dx \\ &= \int_{-2}^1 -x^3 -x^2 + 2x dx \\ &= - \frac{x^4}{4} - \frac{x^3}{3} + x^2 |_{-2}^1 \\ &= \frac{15}{4} - \frac{9}{3} - 3 \\ &= \frac{15}{4} - 6 \\ &= \frac{-9}{4} \end{aligned}

\begin{aligned} M_x &= \int_{-2}^1 \frac{ (1-x^2)^2 - (x-1)^2 }{2} dx \\ &= \frac{1}{2} \int_{-2}^1 1 - 2x^2 + x^4 - ( x^2 - 2x + 1 ) dx \\ &= \frac{1}{2} \int_{-2}^1 -3x^2 +2x + x^4 dx \\ &= \frac{1}{2} ( -x^3 + x^2 + \frac{1}{5} x^5 |_{-2}^1) \\ &= \frac{1}{2} ( \frac{1}{5} - 12 + \frac{32}{5} ) \\ &= \frac{1}{2} ( \frac{33}{5} - \frac{60}{5} ) \\ &= \frac{-27}{10} \end{aligned}

3. Now we solve for x and y

We can use these equations to solve for physical problems (not to be confused with real-world applications), like when your project manager gives you this question:

Q: From x = \frac{\pi}{2} to x = \pi, the density of a rod is given by \rho (x) = \sin x. Find the total mass of the rod.

A:

Q: If \rho (x) = \sqrt{x} is radial density of a disk of radius 4, find its mass.

\text{Work $=$ Force $\times$ Distance} \\~\\ f = ma

Note: Hooke’s law says that the force required to stretch a spring from equilibrium is proportional to distance from equilibrium x.

Q: It takes 10N of force in negative direction to compress a spring 0.2m from equilibrium.

• How much work is done to stretch spring 0.5 m from equilibrium?

A: \begin{aligned} F &= 10 \\ x &= -0.2 \end{aligned} \\~\\ F = kx \\ -10 = k (-0.2) \\ k = \frac{10}{0.2} = 50

• Thus, F = 50x

\begin{aligned} W &= \int_0^{.5} 50 x dx \\ &= 25 x^2 |_0^{.5} \\ &= \frac{25}{4} \end{aligned}

• Or, 6.25 Joules of work

Q: Suppose a cylindrical tank of radius 4m and height 10m is filled with water to a height of 8m,

• How much work is done to pump water out of tank?
  • Assume was raise all water to top.
• Note: Water has weight density of 9800 N/m^3

You can derive other hyperbolic functions trivially, e.g., \sech x = \frac{1}{\cosh x} \\ \cosh^2 x + \sinh^2 x = 1

\text{\textbf{Sequence}: An ordered infinite list of numbers.} \\ \small\text{(often of the form $a_1, a_2, a_3, a_4, ..., a_n, ...$)}

Arithmetic Sequence: Distance between every pair of successive terms is same.

• This difference is called the common difference (d)

Geometric Sequence: Ratio of successive to previous term is same.

• This ratio is called the common ratio (r)

\text{Shorthand for Sequence:} \\ \{ a_n \}_{n=1}^\infin \quad \text{or} \quad \{ a_n \}

Example: Shorthand representation of a sequence

• Listing: { 2, 4, 8, 16, 32, 64, … }
• Formula: \{ 2^n \}

Terminology:

• For a sequence a_1, a_2, a_3, a_4, ..., a_n, ...:
  • n is the index variable.
  • a_n is the term.
  • a_n = 2^n is an example of an explicit function for a sequence
• Sometimes sequences are defined by recurrence relations, where one or more terms are defined explicitly while the rest are defined in terms of previous terms.
  • E.g., a_n = 2^n could also be defined with a recurrence relation where:
    • a_1 = 2
    • a_n = 2a_{n-1} for n \gt 1

\text{A) } - \frac{1}{2} , \frac{2}{3}, -\frac{3}{4}, \frac{4}{5}, - \frac{5}{6}

For the moment ignoring the negatives, we can see that the number is always the index variable, while the denominator is the index variable + 1. Thus, the current formula is:

a_n = \frac{n}{n+1} \times ???

To get the alternating signs we can use { (-1)^n }, which has a pattern of { -1, 1, -1, 1, …}

Note: To alternate on every other term, do { (-1)^{n+1} }

Thus, the explicit formula is:

a_n = \frac{(-1)^n \times n}{n+1}

\text{B) } \frac{3}{4}, \frac{9}{7}, \frac{27}{10}, \frac{81}{13}, \frac{243}{16}, ...

We can see that the numerator is 3^n

• (This is a geometric sequence)

The denominator is 1+3n

• (This is an arithmetic sequence)

Q: Find the explicit formula for 25, 21, 17, 13, 4, 5, 1, -3, …

Q: Find the explicit formula for 12, 18, 27, 40.5, …

Q: Find the explicit formula for: a_1 = 2 \qquad a_n = -3 a_{n-1} \text{ for } n \ge 2

A: By listing out the terms we can see that the sequence is geometric, where r = -3: \{ 2, -6, 18, -54 \}

Therefore, a_n = 2(-3)^{n-1}

Q: a_1 = \frac{1}{2} \qquad a_n = a_{n-1} + ( \frac{1}{2} )^n \text{ for } n \ge 2

A: Listing out the terms we see: \{ \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16} \}

• The denominator is a geometric sequence: 2^n
• And the numerator is the denominator - 1, so:

To graph a sequence { a_n }, we plot points (n, a_n)

Example: { 2^n }

• Points would be (1,2), (2,4), (3,8), …

Given a sequence { a_n }, if the terms a_n get arbitrarily close to some finite number L as n becomes sufficiently large, then we say { a_n } is a convergent sequence is a convergent sequence and that L is the limit of the sequence: \lim_{n \to \infin} a_n = L

Otherwise, we say it is divergent.

Example (as n \to \infin)

• Convergent: 2^n \to \infin
• Convergent: 1 - \frac{1}{2^n} \to 1
• Divergent: (-1)^n \to oscillates between -1 and 1 forever
• Convergent: \frac{(-1)^n}{n} \to 0

Q: \lim_{n \to \infin} \frac{3n^4 = 7 n^2 + 5}{6-4 n^4}

A: We’ll multiply numerator and denominator by \frac{1}{n^4} (so that we can cancel things out with the fact that \frac{1}{x} = 0 if x \to \infin) \begin{aligned} =& \lim_{n \to \infin} \frac{ 3- \frac{7}{n^2} + \frac{5}{n^4} }{ \frac{6}{n^4} - 4 } \\ =& \frac{3 - 0 + 0}{0 - 4} \\ =& - \frac{3}{4} \end{aligned}

Q: \{ (1 + \frac{4}{n})^n \}

Note: 1^\infin is indeterminate

• (We don’t know if we are approaching 1 fast enough to cancel out the infinity)

A: We’ll take the \ln of both sides to get a real value, and then plug that value back in and put both sides as a power of e:

\ln y = \ln (1 + \frac{4}{x})^x \\ = \lim_{x \to \infin} \frac{1 + \frac{4}{x}}{\frac{1}{x}}

This form is indeterminate, so we can use L’Hopitals:

\lim{x \to \infin} \frac{ \frac{-4x^2}{1 + \frac{4}{x}} }{ -x^{-2} } = \lim_{x \to \infin} \frac{4}{1 + \frac{4}{x}} = 4

Q: { \frac{(-1)^n \cos n}{n^2} }

A:

-1 \le \cos n \le 1 \\ \frac{-1}{n^2} \le \frac{\cos n}{n^2} \le \frac{1}{n^2} \\

\frac{-1}{n^2} \land \frac{1}{n^2} \text{ for go to $\infin$ as $n \to \infin$}

• So \frac{\cos n}{n^2} will also go to zero.

Thus, { \frac{(-1)^n \cos n}{n^2} } also converges on zero.

Q: Show r^n converges for -1, r \le 1 and diverges otherwise.

A:

- | r |^n \le r^n \le |r|^n \\ \lim_{n \to \infin} \pm |r|^n = 0

• By squeeze theorem, \lim_{n \to \infin} r^n = 0 when r=1.

• { \frac{1}{n} } is bounded above since \frac{1}{n} \le 1 for all n; and it is also bounded below since \frac{1}{n} \ge 0 for every n.
• { n } is bounded below since n \ge 0 for every n, but it isn’t bounded above (it just goes to infinity). Thus, it is unbounded.

Monotonic Sequences

A sequence { a_n } is monotonic for all n>n_0 if it is either increasing or decreasing for all n>n_0.

Take the definitions of increasing/decreasing monotonic sequences:

1. Subtract a_n from both sides to get the difference test for increasing/decreasing monotonicity.
2. Divide a_n from both sides to get the ratio test for increasing/decreasing monotonicity.
   • Remember that the ratio test should only be used for positive sequences.

Q: Show that { \frac{n+1}{n} } is monotonic.

A: Method 1: Difference Method

We can take the definition of increasing and decreasing monotonic sequences and move the terms to one side to get these formulas:

\text{Increasing Monotonic: } a_{n+1} - a_n \ge 0

\text{Decreasing Monotonic: } a_{n+1} - a_n \le 0

Now we can find a_n and a_{n+1}: a_n = \frac{n+1}{n} \\ a_{n + 1} = \frac{n+2}{n + 1} \\

Now we just need to perform a_{n+1} - a_n and check whether it is negative

• (\le \lor \ge 0)

a_{n+1} - a_n = \frac{(n+2)n}{(n+1)n} - \frac{(n+1)(n+1)}{(n+1)n} \\ = \frac{ n^2 + 2n - ( n^2 -n + n + 1 ) }{ (n+ 1) n } \\ = \frac{ -1 }{ (n+1)n }

Q: Show that { \frac{n+1}{n} } converges.

A: 1. Using Monotone Convergence

In the previous example—which used the difference method—we know that the sequence is decreasing monotonic.

Since n is positive, \frac{n+1}{n} is always positive; or \frac{n+1}{n} > 0, which means that the sequence is bounded from below.

We can calculate that a_1 = 2, and since the sequence is decreasing monotonic, the sequence is also bounded from above.

Therefore, the sequence is convergent by the Monotone Convergence Theorem.

A: 2. L’Hopital’s Rule

We could also take L’Hopital’s and find L=1, as this example was easy to computer.

• But not examples will be like this!

Q: Show that { \frac{4^n}{n!} } converges.

A:

We can see that \lim_{n \to \infin} n! = \infin, which means that \lim_{n \to \infin} \frac{4^n}{n!} = \frac{\infin}{\infin}.

• As this isn’t a function, we can’t use L’Hopital’s, though.

1. Show { \frac{4^n}{n!} } is monotonic

Method 2: Ratio Method

Instead of moving terms in the definition of monotonic sequences with subtraction to create a test, we could also use division to create a test for monotonicity.

• Note: This only works if a_n > 0

\text{Increasing Monotonic: } \frac{a_{n+1}}{a_n} \ge 1

\text{Decreasing Monotonic: } \frac{a_{n+1}}{a_n} \le 1

a_n = \frac{4^n}{n!} \\ a_{n+1} = \frac{4^{n+1}}{(n + 1)!} \\

Doing the test: \begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{4^{n+1}}{(n + 1)!} \times \frac{n!}{4} \\ &= \frac{4n!}{(n+1)!} \\ &= \frac{4n!}{(n+1)n!} \\ &= \frac{4}{n+1} \end{aligned}

Why?

• The 4^{n+1} and \frac{1}{4^n} cancel out into a 4
• In regard to the division of the terms with factorial: \begin{aligned} 3! &= 3 \times 2 \times 1 \\ (3 + 1)! &= 4! = 4 \times 3 \times 2 \times 1 \\ \frac{3!}{(3+1)!} &= \frac{1}{4} \\ \frac{n!}{(n+1)!} &= \frac{1}{n + 1} \\ \end{aligned}

\frac{4}{n+1} will become \le 1 if and only if 4 \le n + 1

• So, for n \ge 3, { \frac{4^n}{n!} } is decreasing.

2. Show { \frac{4^n}{n!} } is bounded

Also, the sequence is > 0 for all n, so the sequence is bounded.

Q: \sum_{n=1}^\infin (\frac{1}{2})^n

Partial Sum of Geometric Series:

Q: \sum_{n=0}^\infin (\frac{-1}{2})^n

A: It converges because -1 < \frac{-1}{2} < 1

If we use \frac{1}{1 - r} we can see that it converges to \frac{2}{3}

Q: \sum_{n=1}^\infin (-1)^n

Q: \sum_{n=1}^\infin \frac{1}{n(n+1)}

A: \begin{aligned} s_1 &= a_1 = \frac{1}{1 \times 2} = \frac{1}{2} \\ s_2 &= a_1 + a_2 = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} = \frac{2}{3} \\ s_3 &= a_1 + a_2 + a_3 = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} = \frac{3}{4} \end{aligned}

• etc…

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4}

• Because \frac{1}{3} > \frac{1}{4}

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... \frac{1}{8} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}

So on and so forth…

Q: Find the sum of: \sum_{n=1}^\infin \ln (\frac{n}{n+1})

A:

1. Let’s find the pattern to find s_k \begin{aligned} s_1 &= \ln(\frac{1}{2}) = \ln 1 - \ln 2 \\ s_2 &= \ln(\frac{1}{2}) + \ln(\frac{2}{3}) = \ln 1 - \ln 2 + \ln 2 - \ln 3 = \ln1 - \ln 3 \\ s_3 &= \ln(\frac{1}{2}) + \ln(\frac{2}{3}) + \ln(\frac{3}{4}) = \ln 1 - \ln 2 + \ln 2 - \ln 3 + \ln 3 - \ln 4 = \ln 1 - \ln 4 \end{aligned}

Thus, s_k = - \ln (k + 1)

2. Thus, the sum is:

\begin{aligned} \sum_{n=1}^\infin \ln (\frac{n}{n+1}) &= \lim_{k \to \infin} s_k \\ &= \lim_{k \to \infin} - \ln (k + 1) \\ &= - \infin \end{aligned}

Q: \sum_{n=1}^\infin \fra{1}{n} - \frac{1}{n+1}

A:

• Writing out the first couple s_k values, we can derive that s_k = \frac{k}{k+1}, which goes to 1 as k goes to infinity.

Q: \sum_{n=1}^\infin \frac{4}{(-3)^n}

Suppose \sum_{n=1}^\infin converges.

• Then the sequence of partial sums (s_k) converges to some number L, where \lim_{k \to \infin} s_k = L

Let’s define a new sequence: b_1 = 0, b_2 = s_1, b_3 = s_2, b_4 = s_3, ..., b_k =s_{k=1}, ...

• Then \sum_{k=1}^\infin b_k = \lim_{k \to \infin} s_{k-1} = L

Then: \ \begin{aligned} lim_{k \to \infin} s_k - b_k &= \lim_{k \to \infin} s_k - s_{k - 1} \\ &= L - L \\ &= 0 \end{aligned}

\begin{aligned} s_1 - b_1 &= s_1 - 0 = s_1 = a_1 \\ s_2 - b_2 &= s_2 - s_1 = a_1 + a_2 -a_1 = a_2 \\ s_3 - b_2 &= s_3 - s_2 = ... = a_3 \\ s_k - b_k &= a_k \end{aligned}

P \to Q = Q' \to P

• If P, then Q
• If not Q, then not P.

If honey’s what you covet, [then] you’ll find that they [heffulumps and woosels] love it.

• Contrapositive: If heffulumps and woosels don’t love honey, then you don’t covet honey.

Convergence: If \sum_{n=1}^\infin a_n converges, then \lim_{n \to \infin = 0}

• Contrapositive: If \lim_{n \to \infin} \ne 0, then \sum_{n=1}^\infin a_n diverges.
  • Remember, this doesn’t say anything about when \lim_{n \to \infin} = 0!

Q: \sum_{n=1}^\infin \frac{n}{3n+1}

A: \lim_{n \to \infin} \frac{n}{3n+1} = \frac{1}{3}

• Thus, diverges. (\frac{1}{3} \ne 0)

Q: \sum_{n=1}^\infin e^{\frac{1}{n^2}}

A: \lim_{n \to \infin} e^{\frac{1}{n^2}} = 1

• Thus, diverges. (1 \ne 0)

Let { a_n } be a sequence of positive terms.

Suppose a continuous and decreasing function f where f(n) = a_n for all n.

Then \sum_{n \to N}^\infin a_n IIF \inf_N^\infin f(x) dx

0 < \sum_{n=2}^\infin a_n \le \int_1^\infin f(x) dx \le \sum_{n=1}^\infin a_n

• If \sum_{n=1}^\infin a_n converges, then \int_1^\infin f(x) dx converges.

Q: \sum_{n=1}^\infin = \frac{1}{e^n}

A:

Let’s test if this integral converges: \begin{aligned} \int_1^\infin \frac{1}{e^x} dx &= \lim_{t \to \infin} \int_1^t e^{-x} dx \\ &= \lim_{t \to \infin} e^{-x} |_1^t \\ &= \lim_{t \to \infin} e^{-t} - e^{-1} \\ &= 0 - \frac{1}{e} \\ &= \frac{1}{e} \\ \end{aligned}

• So, by the integral test, the series converges.

\sum_{n=1}^\infin \frac{1}{n}

\begin{aligned} \int_1^\int \frac{1}{x} dx &= \lim_{t \to \infin} \ln x |_1^t \\ &= \lim_{t \to \infin} \ln t \\ &= \infin \end{aligned}

• By the integral test, the harmonic series diverges.

We can use the integral test to determine if a p-series converges or diverges:

\begin{aligned} \int_1^\infin \frac{1}{x^P} dx &= \lim_{t \to \infin} \int_1^t x^{-P} dx \\ &= \lim_{t \to \infin} \frac{x^{-P-1}}{-p + 1} |_1^t \\ &= \lim_{t \to \infin} \frac{1}{(1-p)x^{p - 1}} |_1^t \\ &= \lim_{t \to \infin} \frac{1}{(1-p)t^{p - 1}} - \frac{1}{1-p} \\~\\ &= 0 - \frac{1}{1-p} \text{ if $p-1 > 0$} \\ &= \infin \text{ if $p-1 < 0$} \\ \end{aligned}

Thus,:

• If p > 1, the series converges.
• If p \le 1, the series diverges.

Q: \frac{1}{n^4}

• A: Converges

Q: \frac{1}{n^{2/3}}

• A: Diverges

Q: \frac{1}{n^\pi}

• A: Converges

Derived from integral test.

S_N + \sum_{n=N+1}^\infin a_N \le S_N + \int_N^\infin f(x) dx

\sum_{n=1}^\infin a_n \le S_N + \int_N^\infin f(x) dx

\int_{N+1}^\infin f(x) dx \le \sum_{n=1}^\infin a_n - S_N \le \int_N^\infin f(x) dx

• R_N = \sum_{n=1}^\infin a_n - S_N

Q: Consider \sum_{n=1}^\infin \frac{1}{n^2}, which is a convergent p-series (because 2 > 1). Calculate S_{10} along with the error.

A: Using a calculator, we get S_{10} \approx 1.549767731

Now, estimating error:

\begin{aligned} \int_{N+1}^\infin \frac{1}{x^2} dx &= \lim_{t \to \infin} \int_N^t x^{-2} dx \\ &= \lim_{t \to \infin} \frac{x^{-1}}{-1} |_N^t \\ &= \lim_{t \to \infin} - \frac{1}{x} |_N^t \\ &= \frac{1}{N} \end{aligned}

\frac{1}{11} < R_{10} < \frac{1}{10} = 0.1

• Thus, the answer S_{10} is at most 0.1 off from the actual answer.

Follow-Up Question: How far would we need to go to get an error at most 0.001?

We now know: R_N < \int_N^\infin \frac{1}{x^2} dx = \frac{1}{N}

Thus we just need to solve: \frac{1}{N} < 0.001

• This gives us N=1000, which is the answer.

\text{Does it converge?: } \sum_{n = 1}^\infin \frac{1}{n^2 + 1}

• Looks kinda like the p-series { \frac{1}{n^2} }
  • And we know that the p-series converges because p=2

We can see that: \sum_{n = 1}^\infin \frac{1}{n^2 + 1} < \sum_{n = 1}^\infin \frac{1}{n^2}

• Because the series is bounded above and is monotonic, we can see that it converges.

Q: Does \sum_{n=1}^\infin \frac{1}{\ln n} converge?

Hint: n > \ln n

A: We’ll do comparison test against \frac{1}{n}

\sum_{n=1}^\infin \frac{1}{n} < \sum_{n=1}^\infin \frac{1}{\ln n}

Q: Does \sum_{n=1}^\infin \frac{1}{\sqrt{n^2 + 3}} converge?

A: We’ll try to directly compare this against \frac{1}{\sqrt{n^2}}, which is the harmonic series:

We know— \sqrt{n^2 + 3} > \sqrt{n^2} —so:

\sum_{n=1}^\infin \frac{1}{\sqrt{n^2 + 3}} < \sum_{n=1}^\infin \frac{1}{n}

• This doesn’t tell us anything, so we’ll use the limit comparison test.
  • (We could also manipulate the inequality to get what we want)

A: Let’s try again with the limit comparison test

\{ a_n \}: \sum_{n=1}^\infin \frac{1}{\sqrt{n^2 + 3}} \qquad \{ b_n \}: \sum_{n=1}^\infin \frac{1}{n}

\begin{aligned} \lim_{n \to \infin} \frac{1}{\sqrt{n^2 + 3}} \times \frac{n}{1} &= \lim_{n \to \infin} \frac{n}{\sqrt{n^2 + 3}} \\ &= \lim_{n \to \infin} \frac{\sqrt{n^2}}{\sqrt{n^2 + 3}} \\ &= \lim_{n \to \infin} \sqrt{ \frac{n^2}{n^2 + 3} } \\ &= \lim_{n \to \infin} \sqrt{ \frac{2n}{2n} } \\ &= \lim_{n \to \infin} \sqrt{ 1 } \\ &= \lim_{n \to \infin} 1 \\ \end{aligned}

• By limit comparison test, we know these both converge or both diverge.
  • Because harmonic series diverges, so does our series.

Q: Does \sum_{n=1}^\infin \frac{1}{3^n - n^2} converge?

A: Limit comparison test

We could compare to { \frac{1}{3^n} } or { \frac{1}{n^2} }.

1. If we try { \frac{1}{n^2} }, we get:

\begin{aligned} \lim_{n \to \infin} \frac{1}{3^n - n^2} \times \frac{n^2}{1} &= \lim_{n \to \infin} \frac{n^2}{3^n - n^2} \\ &= 0 \end{aligned}

• Because the limit goes to zero and { \frac{1}{n^p} } converges, the series converges.

Sometimes, finding the right { b_n } can be difficult.

For example: For \sum_{n=1}^\infin \frac{\ln n}{n^2}, trying to compare against \sum_{n=1}^\infin \frac{1}{n^2} or \sum_{n=1}^\infin \frac{1}{n^2} is inconclusive, we go too small and too big.

• This, however clues us in on trying \sum_{n=1}^\infin \frac{1}{n^{1.5}}, which does work.

For things in trig functions, comparing against the function parameter is usually helpful.

• ex: Compare \sin \theta against \theta, \sin \frac{1}{n} against \frac{1}{n}, etc.

In the previous sections, all tests (except for divergence test) assumed that { a_n } consisted only of positive terms.

\sum_{n=1}^\infin (-1)^{n+1} \frac{1}{n}

The sum of a convergent alternating series lies in-between successive partial sums: S_N < S < S_{N+1}

• Or similar.

Q: Estimate \sum_{n=1}^\infin (-1)^{n+1} \frac{1}{n} within 0.001

A: R_N \le b_{N+1} \le 0.01 \\ R_N \le \frac{1}{N+1} \le 0.01 \\

\begin{aligned} \frac{1}{N+1} &\le 0.01 \\ 1 &\le 0.01(N+1) \\ 100 &\le N+1 \\ 99 &\le N \\ \end{aligned}

A sequence being alternating or having a mix of positive and negative terms makes it corresponding series more likely to converge.

Ex: \sum_{n=1}^\infin (-1)^{n+1} \frac{1}{n} converges, but \sum_{n=1}^\infin \frac{1}{n} diverges.

Also, note that if a series of positive terms converges, then changing the signs of some terms will change what it converges to, but it won’t change the convergence.

Absolute Convergence Theorem Proof:

0 \le a_n + |a_n| \le 2 | a_n |

By assumption and algebraic laws for series, \sum_{n=1}^\infin 2 | a_n | = 2 \sum_{n=1}^\infin | a_n |

By the direct comparison test, \sum_{n=1}^\infin a_n + |a_n| also converges

Let’s do subtraction: \sum_{n=1}^\infin a_n + |a_n| - \sum_{n=1}^\infin |a_n| \\ = \sum_{n=1}^\infin a_n \\

• Which is convergent by algebraic laws for series.
  • (Because we subtracted two convergent series to get another convergent series)

Q: \sum_{n=1}^\infin (-1)^{n+1} \frac{1}{n^2}

A:

Let’s take the absolute value to get:

\sum_{n=1}^\infin \frac{1}{n^2}

We can see that this new series is a convergent p-series (because 2 > 1).

• This tells us that our original series converges absolutely.

Q: \sum_{n=1}^\infin (-1)^{n+1} \frac{\sin n}{n^2}

A:

Let’s take the absolute value to get:

\sum_{n=1}^\infin \frac{\sin n}{n^2}

Let’s do comparison test, specifically the direct comparison (because limit comparison wouldn’t work with \sin n):

Letting {b_n} = \frac{1}{n^2}, we can see that the series converges by direct comparison.

• Thus the original sum converges absolutely.

Q: \sum_{n=1}^\infin (-1)^{n+1} \frac{3n + 4}{2n^2 + 3n + 5}

A:

Let’s take the absolute value to get:

\sum_{n=1}^\infin \frac{3n + 4}{2n^2 + 3n + 5}

We’ll compare this against { \frac{n}{n^2} } (limit comparison test):

\begin{aligned} \lim_{n \to \infin} \frac{3n + 4}{2n^2 + 3n + 5} \times \frac{n}{1} &= \lim_{n \to \infin} \frac{3n^2 + 4n}{2n^2 + 3n + 5} \\ &\textit{(do l'hopitals)} &= \lim_{n \to \infin} \frac{6}{4} \\ \end{aligned}

• This is the case where a_n and b_n either both converge or diverge.
  • As b_n is harmonic series, this tells us that the original series does not converge absolutely.
    • The original series could still be conditionally convergent or divergent, so we don’t have an answer yet.

Doing the alternating series test on the original series, we can see that it is convergent (\lim_{n \to \infin} \frac{3n + 4}{2n^2 + 3n + 5} = 0).

• This combined with the fact that it isn’t absolutely convergent means that the original series converges conditionally.

Q:

\sum_{n=1}^\infin (\frac{3n+1}{5n-2})^n

A:

\begin{aligned} \lim_{n \to \infin} n \sqrt{ (\frac{3n+1}{5n-2})^n } &= \lim_{n \to \infin} \frac{3n+1}{5n-2} \\ &= \frac{3}{5} \end{aligned}

Recall:

\sum_{n=0}^\infin r^n = \frac{1}{1-r}

• for -1 < r < 1
  • diverges for |r| \ge 1
• We can think of this as a function with r=x our input:

f(x) = \sum_{n=0}^\infin x^n = 1 + x + x^2 + x^3 + ... = \frac{1}{1-x}

• for -1 < x < 1

Let’s use the Ratio Test to determine if this power series converges:

Q: \sum_{n=1}^\infin \frac{x^n}{n}

A:

\begin{aligned} \lim_{n \to \infin} | \frac{a_{n+1}}{a_n} | &= \lim_{n \to \infin} | \frac{x^{n+1}}{n+1} \times \frac{n}{x^n} | \\ &= \lim_{n \to \infin} |x| \times \frac{n}{n+1} \\ &= |x| \lim_{n \to \infin} \frac{n}{n+1} \\ &= |x| \end{aligned}

Thus, the outcome of the ratio test depends on what |x| is

• If |x|<1: Converges
• If |x|>1: Diverges
• If |x|=1: Inconclusive

Investigating the Inconclusive Case:

If |x|=1, then there are two cases: x=1 and x=-1

Case I: x=1 \sum_{n=1}^\infin = \frac{1^n}{n}

• Diverges (harmonic series)

Case II: x=-1 \sum_{n=1}^\infin = \frac{(-1)^n}{n}

• Converges (alternating harmonic series)

Conclusion:

• The power series converges when |x|<1 or x=-1
  • Diverges for all other values
  • -1 \le x < 1
• Alternative phrasing: Converges for x = [-1, 1)
  • This is why we need to use other tests to check if a power series converges.

Definitions

Radius of Convergence (R): Distance from the center of the series you can be before it diverges.

Interval of Convergence: Interval on which power series converges.

The Idea: Use the ratio test on power series

• Let’s do this more generally

\lim_{n \to \infin} | \frac{c_{n+1} (x-a)^{n+1}}{ c_n (x-a)^n } | = | x - a | \lim_{n \to \infin} | \frac{c_{n+1}}{c_n} | \\ = |x-a| \rho

Cases:

1. \rho = \infin: Diverges for all x except x=a
2. \rho = 0: Converges for all x
3. \rho = \infin: Converges as long as |x-a| < R
   • R is another name for \frac{1}{\rho}, the Radius of Convergence if the series.

Q: \sum_{n=1}^\infin \frac{(x-1)^n}{n^2}

A: Let’s use the ratio test

\lim_{n \to \infin} | \frac{ (x-1)^{n+1} }{ (x+1)^2 } \times \frac{n^2}{(x-1)^n} | \\ = |x-1| \lim_{n \to \infin} \frac{n^2}{(n+1)^2} \\ = |x-1| < 1

• Radius of Convergence is 1

Q:

\sum_{n=0}^\infin n! (x-3)^n

A: Let’s do ratio test

Recall: \boxed{ \begin{aligned} &\qquad \text{Ratio Test:} \\ &\text{Let $\sum_{n=1}^\infin a_n$ have nonzero terms} \\ &\text{Suppose $\lim_{n \to \infin} | \frac{a_{n+1}}{a_n} | = \rho$} \\ &\text{1. If $\rho < 1$, the series converges absolutely} \\ &\text{2. If $\rho > 1$, the series diverges} \\ &\text{3. If $\rho = 1$, test is inconclusive} \end{aligned} }

\begin{aligned} \lim_{n \to \infin} | \frac{ a_{n+1} }{ a_n } | &= \lim_{n \to \infin} | \frac{ (n+1)! (x-3)^{n+1} }{ n! (x-3)^n } | \\ &= \lim_{n \to \infin} (n+1) |x-3| \\ &= |x-3| \lim_{n \to \infin} (n+1) \\ &= \infin \end{aligned}

• Thus, the series converges at 3 (its center) and nowhere else.
  • Remember: The series always converges at its center.
• Radius of Convergence: 0
  • (You can’t go any distance from 3 without diverging)
• There isn’t really an interval to converge on.

Tip: Manipulating Factorials for Cancellation (n+1)! = (n+1)n!

Q:

\sum_{n=1}^\infin ( \frac{ 6x-12 }{ n } )^n

A:

Thinking:

• We could use the ratio test, but the root test will work better.

Recall:

\boxed{ \begin{aligned} &\qquad \text{Root Test:} \\ &\text{Let $\sum_{n=1}^\infin a_n$ be a series} \\ &\text{Suppose $\lim_{n \to \infin} \hphantom{}^n \sqrt{|a_n|} = \lim_{n \to \infin} | a_n |^{\frac{1}{n}} = \rho$} \\ &\text{1. If $\rho < 1$, the series converges absolutely} \\ &\text{2. If $\rho > 1$, the series diverges} \\ &\text{3. If $\rho = 1$, test is inconclusive} \end{aligned} }

\begin{aligned} \lim_{n \to \infin} \hphantom{}^n \sqrt{ | \frac{ 6x-12 }{ n } |^n } &= \lim_{n \to \infin} \frac{ |6x-12| }{ n } \\ &= |6x-12| \lim_{n \to \infin} \frac{ 1 }{ n } \\ &= 0 \end{aligned}

• This is zero (<1) for every x, so the series always converges.
• Radius of Convergence: Infinity
• Interval of Convergence: (-\infin, \infin)

Q:

\sum_{n \to \infin}^\infin \frac{ (-2x+4)^n }{ n^2 \times 8^n }

A: Ratio Test

\begin{aligned} \lim_{n \to \infin} | \frac{(-2x+4)^{n+1}}{(n+1)^2 \times 8^{n+1}} \times \frac{ n^2 \times 8^n }{ (-2x+4)^n } | &= \frac{ |-2x+4| }{ 8 } \lim_{n \to \infin} \frac{n^2}{(n+1)^2} \\ &= \frac{ |-2x+4| }{ 8 } \times 1 \end{aligned}

This series will converge when \frac{ |-2x+4| }{ 8 } < 1, so now we need to solve for that.

\begin{aligned} \frac{ |-2x+4| }{ 8 } &< 1 \\ |-2x+4| &< 8 \\ 2 \times | x-2 | &< 8 \\ | x-2 | &< 4 \\ | x-2 | &< 4 \\ \end{aligned}

• Radius of Convergence: 4
• We know our center is 2 and R=4, thus the work-in-progress interval is -2, 6
  • However, at the endpoints the ratio test is inconclusive (equal to 1)
    • Thus we need to test the endpoints to know what the interval of convergence is.

Aside: Understanding | x-2 | < 4

• “x’s distance from 2 is at most 4”

Testing Endpoints:

Case I: -2 \sum_{n=1}^\infin \frac{ (-2(-2) + 4)^n }{ n^2 \times 8^n } = blah

• Converges because it yields the convergent p-series \frac{1}{n^2}

Case II: 6

• Doing a bunch of algebra we eventually get— \sum_{n=1}^\infin (-1)^n \frac{1}{n^2}
• —which converges by the alternating series test.

\begin{aligned} \frac{2}{(x-1)(x-3)} &= \frac{-2}{1 - x} \times \frac{-1}{3 - x} \\ &= \frac{2}{1-x} \times \frac{\frac{1}{3}}{1 - \frac{x}{3}} \\ &= (\sum_{n=0}^\infin 2x^n)(\sum_{n=0}^\infin \frac{1}{3} (\frac{x}{3})^n) \\ &= (2 + 2x + 2x^2 + ...)(\frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + ...) \end{aligned}

e_0 = c_0 d_0 = 2(\frac{1}{3}) = \frac{2}{3} \\ e_1 = c_0 d_1 + c_1 d_0 = 2 ( \frac{1}{4} ) + 2 ( \frac{1}{3} ) \\