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$$ \sum_{k=1}^{\infty} \frac{e^k}{k^k} $$

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!

Somos
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  • I wouldn't expect this to have a pleasant closed form, numerical methods are likely to be best. – lulu Jan 07 '21 at 14:57
  • You wrote without using a program. Can you however use an existing approximation of $e$ or not? – mathcounterexamples.net Jan 07 '21 at 14:58
  • @lulu This is very interesting to me how people know whether a sum or an integral has a closed form, is there a way of knowing or at least guessing with high "probability" that an integral or a sum does not have a nice closed form? I would like to know (really, this is very interesting) –  Jan 07 '21 at 14:58
  • I don’t think it can be evaluated per se, but you can estimate it to any degree of accuracy. – Clayton Jan 07 '21 at 14:59
  • Can you compute $a_{k+1}/a_k$ where $a_k = e^k/k^k$ and determine the behaviour? –  Jan 07 '21 at 14:59
  • All methods are valid unless you write a Python/C/ etc.. program with a for loop that counts until a 1,000~ and give the answer :-) I really tried my best to solve this, many hours and I even did write a program just to know what is the answer –  Jan 07 '21 at 15:01
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    @JetPlane Nothing much to it, I am afraid. I've seen a lot of analytic expansions and over time you get a sense of how new expressions fit in. Hardly a foolproof method, of course. Helps to realize that simple analytic expressions are few and far between, so any random looking thing one writes down (involving infinite operations like sums or integrals) is unlikely to have a pleasant simplification. – lulu Jan 07 '21 at 15:13
  • Related: https://math.stackexchange.com/questions/115410/whats-the-sum-of-sum-limits-k-1-infty-fractkkk – Hans Lundmark Jan 07 '21 at 16:09

2 Answers2

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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 $e= \sum_{k=0}^\infty \frac{1}{n!}$.

mathcounterexamples.net
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  • I don't think that $a_8$ is an approximation of $a$; in fact, you've shown that $a_8<10^{-4}$. Guess you've meant $a-a_8$ instead. – lisyarus Jan 07 '21 at 15:37
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    Thanks for noticing a typo. In fact it is my initial definition of $a_n$ which was wrong. But $a_n$ is indeed the approximation of $a$. – mathcounterexamples.net Jan 07 '21 at 15:39
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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 :-)

  • Seems reasonable, but other programs say it actually converges to about 5~ what happened there? –  Jan 07 '21 at 15:02
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    $\frac{e^k}{k^k} = (\frac{e}{k})^k < (\frac{1}{2})^k$ for $k\ge6$, so it is bounded above by a converging geometric series. – Jaap Scherphuis Jan 07 '21 at 15:10
  • @JaapScherphuis Thank you Jaap, I had not given it enough thought. I have cited your comment into my 'answer'. Thanks again. –  Jan 07 '21 at 15:14
  • Your argument merely shows that, after the division by $k^k$, the $k$-th term contributes at least $1/k!$. So you've bounded it below by a series, but unfortunately for your argument that series converges so fails to prove divergence of the original series. – Jaap Scherphuis Jan 07 '21 at 15:18