$$ \sum_{k=1}^{\infty} \frac{e^k}{k^k} $$
The solution is about $\approx {5.5804}$
But I don't know how to calculate this sum, I tried using the squeeze theorem but I couldn't find $2$ series that converge such that $$a_n \leq \frac{e^k}{k^k} \leq b_n$$
The only thing I did is set $a_n = \frac{1}{k^k}$ but it didn't really help because it's way less than the original question's series, by a couple of magnitudes... ($e^k$ to be exact)
I would highly appreciate your help :-) Thanks!
On
Denote $$ a=\sum_{k=1}^{\infty} \frac{e^k}{k^k}, \, a_n=\sum_{k=1}^n \frac{e^k}{k^k}.$$
We have for $n \in \mathbb N$ $$0 \le a - a_n = \sum_{k=n+1}^\infty \frac{e^k}{k^k} \le \sum_{k=n+1}^\infty \frac{e^k}{(n+1)^k} = \left(\frac{e}{n+1}\right)^{n+1}\frac{n+1}{n+1-e}.$$
If I understand well the question, you expect to compute an approximation of $a$ at $10^{-4}$. Therefore you need
$$\left(\frac{e}{n+1}\right)^{n+1}\frac{n+1}{n+1-e} \lt 10^{-4}$$ which is the case for $n=8$.
$a_8$ is an approximation of $a$ at $10^{-4}$ and you have
$$a_8 \approx 5.580377.$$ With the same process, you can compute the value of $a$ (with a paper, a pencil and some sweat...) at the desired precision.
Note: this supposes that an approximation of $e$ is already known. If not, you can find one using $e= \sum_{k=0}^\infty \frac{1}{n!}$.
If you expand out $e^k$ you should find term in $k^k/k!$ with all terms positive, so I shouldn't think it will converge.
Edit
I hadn't considered Japp's thought below, seems I could be wrong, you have no idea how hard it is to type this :-)