# Do the following series converge? [closed]

Say we have a $a_n$, a sequence of numbers $a_n = \sum_{k=1}^{n} e^{-k}$.

a) Does the $\lim_{n\to\infty}a_n$ exist?

b) Does the series converge: $\sum_{n=1}^\infty a_n$?

c) Does the series converge: $\sum_{n=1}^\infty \frac{a_n}{n^2}$?

d) Does the series converge: $\sum_{n=1}^\infty \frac{a_n}{n}$?

e) Does the series converge: $\sum_{n=1}^\infty \frac{a_n}{n+n^2}$?

can anyone help me with this problem? I'm not sure how to answer these

## closed as off-topic by user223391, Thomas, Winther, uranix, GEdgarSep 12 '15 at 13:47

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We have $$\lim_{n\to+\infty}a_n=\sum_{k=1}^{+\infty}\mathrm{e}^{-k}=\sum_{k=1}^{+\infty}\left(\frac{1}{\mathrm{e}}\right)^{k}=\frac{\frac{1}{\mathrm{e}}}{1-\frac{1}{\mathrm{e}}}=\frac{1}{\mathrm{e}-1}$$ which is a well known result about the geometric series (since $\frac{1}{\mathrm{e}}<1$). From here, you should be able to prove the convergence of the other series by comparison test.

a) The sum can be calculated by the geometric series.

$$a_n = q +q^2+q^3+...+q^n=q\frac{1-q^{n}}{1-q}$$ where $q = \frac{1}{e}$.

You will get $$a=\lim_{n\rightarrow \infty}a_n=\frac{1}{e}\frac{1}{1-\frac{1}{e}}=\frac{1}{e-1}$$

b)You can check the necessary condition for convergence here $$\lim_{n \rightarrow \infty}a_n=0$$

c) Find a upper bound for the series e.g. $\frac{a_n}{n^2}<\frac{1}{e-1}\frac{1}{n^2}$.

d) Find a lower bound for the series and show that it diverges, hence the series does diverge.

e) Similar to c but now with additional estimate on $\frac{1}{n+n^2}<\frac{1}{n^2}$

For (a) use the fact that $\lim _{n\to \infty} a_{n}$ is a geometric series.

For (b) recall that for an infinite series $\sum_{n=0}^{\infty} a_{n}$ to converge it must be that $a_{n} \to 0$.

For (c) try the ratio test.

For (d) try the comparison test with $\sum_{n=0}^{\infty} \frac{1}{n}$

For (e) try comparison test with $\sum_{n=0}^{\infty} \frac{a_{n}}{n^{2}}$ (i.e. part (c))