Determine if $\displaystyle \sum_{n=3}^\infty \frac{(-1)^{n+1}}{e^n}$ converges absolutely, converges conditionally, or diverges.
What do the absolute value of the terms of this series look like?
After taking the absolute value, we get the series $\displaystyle \sum_{n=3}^\infty \frac{1}{e^n}$. How can we determine if this series converges?
Try the integral test on this series.
$\displaystyle \sum_{n=3}^\infty \frac{(-1)^{n+1}}{e^n}$ converges absolutely.
Determine if $\displaystyle \sum_{n=2}^\infty \frac{(-1)^{n}}{n^{8/9}}$ converges absolutely, converges conditionally, or diverges.
What do the absolute value of the terms of this series look like?
After taking the absolute value, we get the series $\displaystyle \sum_{n=3}^\infty \frac{1}{n^{8/9}}$. How can we determine if this series converges?
By $p$-series, the series of absolute values diverges. What about the original series?
Does the alternating series test apply to the original series? If so, what does that mean about its convergence?
$\displaystyle \sum_{n=2}^\infty \frac{(-1)^{n}}{n^{8/9}}$ converges conditionally.
Determine if $\displaystyle \sum_{n=2}^\infty \frac{\sin\left(\frac{n\pi}{7}\right)n}{\sqrt{n^2 + 1}}$ converges absolutely, converges conditionally, or diverges.
What happens to the terms of this series as $n \rightarrow \infty$? What happens if you ignore the sine term?
Without the sine part, the terms go to 1. What does that mean if you factor in the sine part?
If you add in the sine part, it means this limit does not exist. In particular, the limit does not equal zero. What does that mean for the series?
$\displaystyle \sum_{n=2}^\infty \frac{\sin\left(\frac{n\pi}{7}\right)n^2}{(n-1)(n+3)}$ diverges by the $n$th term divergence test.
Determine if $\displaystyle \sum_{n=4}^\infty \frac{(-1)^n}{n + \frac{1}{n}}$ converges absolutely, converges conditionally, or diverges.
What do the absolute value of the terms of this series look like?
After taking the absolute value, we get the series $\displaystyle \sum_{n=3}^\infty \frac{1}{n + \frac{1}{n}}$. How can we determine if this series converges?
Use the fact that for $n \geq 4$, $\frac{1}{n} \leq n$ to say that $\frac{1}{n + \frac{1}{n}} \geq \frac{1}{n + n} = \frac{1}{2n}$. What do we know about this series?
Since that diverges, direct comparison says that the absolute value series diverges. Does the Alternating Series test apply to this series? What does that mean?
$\displaystyle \sum_{n=4}^\infty \frac{(-1)^n}{n + \frac{1}{n}}$ converges conditionally using the alternating series test.
How many terms of the series $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ do I need to approximate the sum of the series with an error at most $0.01$?
This is an alternating series, so we can use the error bound for alternating series to work this out. What does this error bound say?
The error bound tells us that $|S_N - S| \leq b_{N+1}$ where $b_{N+1}$ is the first term that we leave off of the series. How does this help us?
To keep the error within $0.01$, we need that $b_{N+1} = \frac{1}{\sqrt{N+1}} \leq 0.01$. We solve this out for $N$
Thus $\sqrt{N+1} \geq 100$, so $N \geq 100^2 - 1$
We need to take $N \geq 9999$.
Determine if $\displaystyle \sum_{n=2}^\infty \frac{(-1)^n}{n + \sqrt{n}}$ converges absolutely, converges conditionally, or diverges.
How many terms do we need to approximate the sum of the series $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ to within $10^{-9}$?