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Exercises relating to various topics we have studied.

The series is convergentdivergent by limit comparison to the $p$-series with $p = \answer {1}$. Likewise, the series is convergentdivergent by limit comparison to the $p$-series with $p = \answer {2}$.
Use a Maclaurin series to determine the dominant behavior of as $n \rightarrow \infty$.
Fill in the blank below with an appropriate constant to make the series absolutely convergent:
Use the Maclaurin series for $\ln (1 + x)$ and substitute $x = 1/n$.
Determine whether the series below converges absolutely, conditionally, or diverges.
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.
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.
Absolute Conditional Diverge
Try limit comparison with the harmonic series.
Determine whether the series below converges absolutely, conditionally, or diverges.
Absolute Conditional Diverge
Do a Taylor expansion of $\ln (1 + x)$ and evaluate at $x = 1/n$.
Compute the sum of the series below.
• We know the series for $-1 < x < 1$.
• This means
• We conclude
Compute the sum of the series below.
Compute the sum of the series below.
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

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

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