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PAGE 1

Sum of Infinite Series

If \( L = \sum_{n=1}^{\infty} \frac{1}{2^n} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \), then \( L = \)

A. 1/3
B. 2/3
C. 1
D. 4/3
E. 5/3

Geometric Series Formula

Geometric series: \( \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \dots = \frac{a}{1-r} \) for \( |r| < 1 \)

  • \( a \) = first term of series
  • \( r \) = common ratio

Step-by-Step Calculation

1. Evaluating the first series:

\[ \sum_{n=1}^{\infty} \frac{1}{2^n} = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1 \]

2. Evaluating the second series:

\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} = \sum_{n=0}^{\infty} \left(-\frac{1}{2}\right)^n = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \dots \]\[ = \frac{1}{1 - (-1/2)} = \frac{1}{3/2} = \frac{2}{3} \]

3. Final Sum:

\[ L = 1 + \frac{2}{3} = \frac{5}{3} \]

PAGE 2

Convergence of p-series

Find all values of \( p \) for which \( \sum_{n=1}^{\infty} \frac{1}{(n^2 + 1)^p} \) converges.

A. \( p > 1 \)
B. \( p \le 1 \)
C. \( p \ge 1 \)
D. \( p > 1/2 \)
E. \( p \le 1/2 \)

The p-series Test

p-series: \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) converges if \( p > 1 \)

Asymptotic Analysis

To determine convergence, we look at the behavior of the terms as \( n \) becomes large:

\[ \sum_{n=1}^{\infty} \frac{1}{(n^2 + 1)^p} \text{ when } n \text{ is large becomes } \sum \frac{1}{(n^2)^p} = \sum \frac{1}{n^{2p}} \]

Using the p-series test on the simplified form:

Converges if \( 2p > 1 \)

\( p > 1/2 \)

PAGE 3

Conditional and Absolute Convergence

Which of the following series converge conditionally?

(I) \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \]
(II) \[ \sum_{n=2}^{\infty} \frac{(-1)^n n}{\ln n} \]
(III) \[ \sum_{n=1}^{\infty} \frac{(-1)^n n}{e^n} \]
  • A. II only.
  • B. I and III only.
  • C. I and II only.
  • D. All three.
  • E. None of them.

Definitions

conditionally: given \( \sum a_k \), \( \sum |a_k| \) diverges but \( \sum a_k \) converges

absolutely: given \( \sum a_k \), both \( \sum a_k \) and \( \sum |a_k| \) converge

Analysis of Series I

I. \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \) converges, because \( \lim_{n \to \infty} \frac{(-1)^n}{n^2} = 0 \) and \( \frac{1}{n^2} \) decreases (alternating series test).

\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n^2} \right| = \sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges, } p > 1 \]

so I converges absolutely, not conditionally

PAGE 4

Analysis of Series II

\[ \sum_{n=2}^{\infty} \frac{(-1)^n n}{\ln n} \]

divergence test: \[ \lim_{n \to \infty} \frac{n}{\ln n} = \lim_{n \to \infty} \frac{1}{\frac{1}{n}} = \lim_{n \to \infty} n \neq 0 \]

does not converge

Analysis of Series III

III. \( \sum_{n=1}^{\infty} \frac{(-1)^n n}{e^n} \) by alt. series test, this converges

\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n n}{e^n} \right| = \sum_{n=1}^{\infty} \frac{n}{e^n} \]

ratio test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \)

if = 1 inconclusive (same as in root test)

\[ \lim_{n \to \infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}} \right| = \lim_{n \to \infty} \frac{n+1}{e^{n+1}} \cdot \frac{e^n}{n} = \lim_{n \to \infty} \frac{n+1}{n} \cdot \frac{e^n}{e^{n+1}} \]\[ = \lim_{n \to \infty} \frac{n+1}{n} \cdot \frac{1}{e} = \frac{1}{e} < 1 \]

conv. abs.

PAGE 5

Power Series Convergence

Find the interval of convergence of the power series \[ \sum_{n=1}^{\infty} \frac{n}{5^n}(x - 2)^n \]

A. \( -5 < x < 5 \)

B. \( 3 < x < 7 \)

C. \( -2 < x < 2 \)

D. \( -3 \le x < 7 \)

E. \( -3 < x < 7 \)

Ratio Test handles this well

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{n+1}{5^{n+1}}(x-2)^{n+1}}{\frac{n}{5^n}(x-2)^n} \right| \]\[ = \lim_{n \to \infty} \left| \frac{n+1}{5^{n+1}} \cdot \frac{5^n}{n} \cdot \frac{(x-2)^{n+1}}{(x-2)^n} \right| = \lim_{n \to \infty} \left( \frac{n+1}{n} \right) \cdot \left( \frac{5^n}{5^{n+1}} \right) \cdot \left| \frac{(x-2)^{n+1}}{(x-2)^n} \right| \]

Note: As \( n \to \infty \), \( \frac{n+1}{n} \to 1 \) and \( \frac{5^n}{5^{n+1}} \to \frac{1}{5} \).

\[ = \lim_{n \to \infty} \left| \frac{(x-2)}{5} \right| < 1 \implies \left| \frac{x-2}{5} \right| < 1 \]

so \( -1 < \frac{x-2}{5} < 1 \)

\( -5 < x - 2 < 5 \)

\( -3 < x < 7 \)

not done!

Check end values (ratio = 1 at ends)

PAGE 6

Checking Endpoints

\[ \sum_{n=1}^{\infty} \frac{n}{5^n}(x-2)^n \]

at \( x = -3 \), series becomes:

\[ \sum_{n=1}^{\infty} \frac{n}{5^n}(-5)^n = \sum_{n=1}^{\infty} (-1)^n n \quad \text{diverges} \]

at \( x = 7 \), series becomes:

\[ \sum_{n=1}^{\infty} \frac{n}{5^n} 5^n = \sum_{n=1}^{\infty} n \quad \text{diverges} \]

Final Interval of Convergence: \( (-3, 7) \) or \( -3 < x < 7 \).

PAGE 7

Use the power series representation of \(\sin x\) to find the first three terms of the Maclaurin series of \(f(x) = x \sin(x^2)\)

A. \(x^3 + \frac{x^7}{3!} + \frac{x^{11}}{5!}\)
B. \(x + \frac{x^3}{3} + \frac{x^5}{5}\)
C. \(x^3 - \frac{x^7}{3!} + \frac{x^{11}}{5!}\)
D. \(x - \frac{x^3}{3} + \frac{x^5}{5}\)
E. \(x^3 - \frac{x^7}{3} + \frac{x^{11}}{5}\)

On exam, we will give:

\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\]

Step-by-Step Solution

First, substitute \(x^2\) into the series for \(\sin x\):

\[\sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \dots = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \dots\]

Then, multiply the entire series by \(x\):

\[x \sin(x^2) = x^3 - \frac{x^7}{3!} + \frac{x^{11}}{5!} - \dots\]
PAGE 8

Problem 16: Limits using Power Series

16. Recall that \(\cos x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}\) for all \(x\), and \(\tan^{-1} x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}\) for \(|x| \le 1\).

Find the limit:

\[\lim_{x \to 0} \frac{\cos(x^3) + \tan^{-1}\left(\frac{x^6}{2}\right) - 1}{x^{12}}\]
A. \(\frac{1}{4}\)
B. \(-\frac{7}{24}\)
C. \(\infty\)
D. \(\frac{1}{24}\)
E. \(-\frac{1}{3}\)
F. 0

Expansion of Terms

\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots\]\[\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots\]

Substituting the specific arguments:

\[\cos(x^3) = 1 - \frac{x^6}{2!} + \frac{x^{12}}{4!} - \frac{x^{18}}{6!} + \dots\]\[\tan^{-1}\left(\frac{x^6}{2}\right) = \frac{x^6}{2} - \frac{(\frac{x^6}{2})^3}{3} + \dots = \frac{x^6}{2} - \frac{x^{18}}{24} + \frac{x^{30}}{5 \cdot 32} - \dots\]

Combining the Numerator

\[\cos(x^3) + \tan^{-1}\left(\frac{x^6}{2}\right) - 1 = \left(1 - \frac{x^6}{2!} + \frac{x^{12}}{4!} - \frac{x^{18}}{6!} + \dots\right) + \left(\frac{x^6}{2} - \frac{x^{18}}{24} + \frac{x^{30}}{5 \cdot 32} - \dots\right) - 1\]

The constant terms \(1\) and \(-1\) cancel, and the \(x^6\) terms \(-\frac{x^6}{2}\) and \(\frac{x^6}{2}\) cancel:

\[= \frac{x^{12}}{4!} - \frac{x^{18}}{6!} - \frac{x^{18}}{24} + \frac{x^{30}}{5 \cdot 32} + \text{bunch of } x^n, n \ge 30\]

Dividing by \(x^{12}\) and taking the limit as \(x \to 0\), only the coefficient of \(x^{12}\) remains:

\[\text{Limit} = \frac{1}{4!} = \frac{1}{24}\]
PAGE 9

Limit Calculation using Taylor Series

\[ \frac{\cos(x^3) + \tan^{-1}(\frac{x^6}{2}) - 1}{x^{12}} \]
\[ = \frac{\frac{x^{12}}{4!} - \frac{x^{18}}{6!} + \dots}{x^{12}} \]

\( x^n \quad n \ge 18 \)

\[ = \frac{1}{4!} + \dots \]

has \( x \)

\[ \lim_{x \to 0} \left( \dots \right) = \frac{1}{4!} = \frac{1}{24} \]
PAGE 10

Polar Coordinates Problem

A point \( P \) has polar coordinates \( (3, \pi/4) \). Which of the following are also polar coordinates of \( P \)?

  • (I) \( (-3, -\pi/4) \)
  • ✓ (II) \( (-3, 5\pi/4) \)
  • ✓ (III) \( (3, -7\pi/4) \)
  • (IV) \( (3, -5\pi/4) \)

A. I and II only.

B. I and III only.

C. I and IV only.

D. II and III only.

E. II and IV only.

Visualizing the Point P

The original point is located at a radius of 3 and an angle of \( \pi/4 \).

Coordinate graph showing point P at (3, pi/4) in the first quadrant.
I.
Graph for option I showing an angle of -pi/4 and a reflected point.
II.
Graph for option II showing an angle of 5pi/4 with a negative radius.
III.
Graph for option III showing a clockwise rotation of -7pi/4.
IV.
Graph for option IV showing an angle of -5pi/4.
PAGE 11

Polar Equations and Plots

Which of the following polar equations describes the plot?

A cardioid graph on a polar coordinate system, symmetric about the horizontal axis, passing through (2,0) and (0,pi).

Observations from the plot:

  • At \(\theta = 0\) or \(2\pi\), \(r = 2\)
  • At \(\theta = \pi\), \(r = 0\)

Analysis of Options

  • A.

    \(r = 2\)

    Circle with radius 2

  • B.

    \(r = 1 + \cos(\theta)\)

    At \(\theta = 0\), \(r = 1 + \cos(0) = 2\)

    At \(\theta = 2\pi\), \(r = 2\)

  • C.

    \(r = 2 \cos(\theta)\)

    At \(\theta = 0\), \(r = 2\)

    At \(\theta = 2\pi\), \(r = 2\)

    At \(\theta = \pi\), \(r = 2(-1) = -2\)

  • D.

    \(r = 1 - 2 \cos(\theta)\)

    At \(\theta = 0\), \(r = -1\)

    At \(\theta = \pi\), \(r = 3\)

  • E.

    \(\theta = \frac{\pi}{4}\)

    Line

    A Cartesian coordinate system showing a dashed line at a 45-degree angle representing theta equals pi over 4.