Exercises: Cumulative

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Exercises relating to various topics we have studied.

The series $\sum _{n=1}^\infty \left ( e^{ \frac {1}{n}} - 1 \right )$ is convergentdivergent by limit comparison to the $p$ -series $\sum _{n=1}^\infty \frac {1}{n^p}$ with $p = \answer {1}$ . Likewise, the series $\sum _{n=1}^\infty \left ( e^{ \frac {1}{n}} - 1 - \frac {1}{n} \right )$ is convergentdivergent by limit comparison to the $p$ -series with $p = \answer {2}$ .

Use a Maclaurin series to determine the dominant behavior of $e^{\frac {1}{n}} - 1$ as $n \rightarrow \infty$ .

Fill in the blank below with an appropriate constant to make the series absolutely convergent: $\sum _{n=1}^\infty \left [ \ln \left ( 1 + \frac {1}{n} \right ) - \frac {\answer {1}}{n} \right ]$

Use the Maclaurin series for $\ln (1 + x)$ and substitute $x = 1/n$ .

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty n \sin \frac {1}{n}$

Absolute Conditional Diverge

Show that the terms do not go to zero as $n \rightarrow \infty$ .

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {1}{n} \sin \frac {1}{n}$

Absolute Conditional Diverge

Use the Maclaurin series for $\sin x$ and evaluate at $x = 1/n$ . Find the dominant term.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {1}{n} \ln \left (2 + \frac {1}{n} \right )$

Absolute Conditional Diverge

Try limit comparison with the harmonic series.

Determine whether the series below converges absolutely, conditionally, or diverges. $\sum _{n=1}^\infty \frac {1}{n} \ln \left (1 + \frac {1}{n} \right )$

Absolute Conditional Diverge

Do a Taylor expansion of $\ln (1 + x)$ and evaluate at $x = 1/n$ .

Compute the sum of the series below. $\sum _{n=0}^\infty \frac {(-1)^n}{2n+1} \frac {1}{3^n}$

We know the series $\sum _{n=0}^\infty \frac {(-1)^n x^{2n+1}}{2n+1} = \answer {\arctan x}$ for $-1 < x < 1$ .
This means $\sum _{n=0}^\infty \frac {(-1)^n x^{2n}}{2n+1} = \answer {\frac {\arctan x}{x}}.$
We conclude $\sum _{n=0}^\infty \frac {(-1)^n}{2n+1} \frac {1}{3^n} = \answer {\frac {\pi \sqrt {3}}{6}}.$

Compute the sum of the series below. $\sum _{n=0}^\infty n 3^{-n} = \answer {\frac {3}{4}}.$

$\frac {1}{1-x} = \sum _{n=0}^\infty x^n \ \ \text { when } -1 < x < 1.$

$\frac {1}{(1-x)^2} = \sum _{n=0}^\infty n x^{n-1} \ \ \text { when } -1 < x < 1.$

Compute the sum of the series below. $\sum _{n=1}^\infty \frac {3}{n(n+1)} = \answer {3}.$

You don’t need Taylor series for this one.

It’s a telescoping series

Sample Exam Questions

If it converges, find the sum of the series $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n \pi ^{2n}}{3^{2n} (2n)!}$ . If the series diverges, explain why.

$\ln 2$ $\ln 3 - \ln 2$ $e^{-2}$ $\displaystyle \frac {1}{2}$ $\displaystyle \frac {2}{e}$ diverges

We recognize the Taylor series for cosine: $\cos x = \sum _{n=0}^\infty \frac {(-1)^n \pi ^{2n}}{3^{2n}(2n)!}$ The series in question is exactly $\sum _{n=0}^\infty \frac {(-1)^n (\pi /3)^{2n}}{(2n)!} = \cos \frac {\pi }{3} = \frac {1}{2}.$

What is the limit of the sequence $\displaystyle \left \{ n^2 \left ( 1 - \cos \frac {1}{n} \right ) \right \}$ ?

$1$ $-1$ $\displaystyle \frac {\sqrt {3}}{2}$ $\displaystyle \frac {1}{2}$ $\displaystyle -\frac {\sqrt {3}}{2}$ diverges

Find the limit of the sequence $a_n = \left \{ n \left [ \ln (n+3) - \ln n \right ] \right \}.$

$0$ $1$ $\ln 3$ $3$ $\infty$ the limit does not exist