# How can I determine the convergence of the following series

So I've been solving convergence tests of series all day and I got stuck on the following three:

1. $\sum_{k=0}^\infty \frac{4+|Cos{k}|}{k^3}$

2. $\sum_{k=0}^\infty \frac{k!}{k^k}$

3. $\sum_{k=0}^\infty \frac{1}{4+2^{-k}}$

For the first one I've tried to compare it to the series $\sum_{k=0}^\infty \frac{4}{k^3}$ which is a "smaller" series since |cosk| is always a number between 0 and 1. However that series seems to be converging since it is a "p" series where p=3 and p>1 and if the "smaller" series converges it doesn't have to mean that the larger series will converge as well.

For the second one I've tried comparing it to the series $\sum_{k=0}^\infty \frac{k!}{k}$ but that didn't work as well and now I'm stuck at all three.

Any kind of help would be appreciated.

For the second series, note that

$$\frac{k!}{k^k}=\frac{k}{k}\,\frac{k-1}{k}\cdots \frac{2}{k}\,\frac 1k<\frac2{k^2}$$

and that the series $\sum_{k=1}^\infty \frac1{k^2}$ converges.

• Since you're new to the site, I wanted to let you know that once you've accrued enough reputation points, you can up vote answers as you see fit. And of course, you can accept an answer now as you see fit. ;-)) Jun 14 '17 at 20:04

Let $a_k$ denote the $k$-th term.

1. $|a_k| \leq \frac{5}{k^3}$.

2. I'd use Stirling's approximation of the factorial.

3. $a_k \geq \frac15$

• Sorry I don't understand your 3rd answer. To what series should I compare the third one? Jun 8 '17 at 19:09
• @DavidMason: $\frac{1}{4 + 2^{-k}} \geq \frac15$ for $k \geq 0$. Jun 8 '17 at 19:10

1) comparison tests

if $\sum_\limits{k=1}^{\infty} a_k \le \sum_\limits{k=1}^{\infty} b_k$ and $\{b\}$ coverges, then $\{a\}$ converges

if $\sum_\limits{k=1}^{\infty} a_k \ge \sum_\limits{k=1}^{\infty} b_k$ and $\{a\}$ diverges, then $\{b\}$ diverges.

Compare to $\sum\frac {5}{k^3}$

2) try the root test.

Update: I was thinking to use Stirling's aproximation for $k!$ However, lets look at the ratio test, instead.

the series converges if $\lim_\limits{k\to\infty}\left(\dfrac {\frac {(k+1)!}{(k+1)^{k+1}}}{\frac {k!}{k^k}}\right)<1$

$\lim_\limits{k\to\infty}\frac {k^{k+1}}{(k+1)^{k+1}}\\ \lim_\limits{k\to\infty}(1-\frac {1}{k+1})^{k+1} = e^{-1} <1$

3) what is $\lim_\limits{k\to\infty} \frac{1}{4+2^{-k}}$?

• 2) How can I try the root rest if the entire fraction is not on a power k only the denominator. 3) I don't know what the limit is. I can't calculate it manually Jun 8 '17 at 19:07
• I was thinking sterling's approximation, but changed my mind have updated using the ratio test. Jun 8 '17 at 19:18