I need to find the sum of the infinite series with the following general term $\frac{2^{nlogn}}{n!}$

I initially started off by getting rid of the log in the the numerator, so that later became $\frac{n^{nlog2}}{n!}$

So I tried my best to split the terms so that the consecutive terms cancel out, but I couldn't find a way. Am I approaching this the wrong way?

  • 1
    $\begingroup$ The sum diverges...$ n^n\geq n!$ even log2 can't reduce the divergence much. (3log2 is almost 1) $\endgroup$ Nov 13, 2020 at 7:00
  • 2
    $\begingroup$ @Anindya: this is false. $n!$ grows like $\exp(n \log n - n)$ and this is enough to overwhelm the numerator because $\log 2 < 1$. $\endgroup$ Nov 13, 2020 at 7:20
  • 2
    $\begingroup$ @Rumplespaceking: as far as I know series like this don't have a closed form. What do you "need" to find this sum for? $\endgroup$ Nov 13, 2020 at 7:21
  • 3
    $\begingroup$ @Rumplespaceking: that is very surprising to me. I would guess that this is a typo of some kind but I can't imagine what the intended question is. $\endgroup$ Nov 13, 2020 at 8:22
  • 1
    $\begingroup$ Are you probably not just being asked to determine if the series converges or not, rather to find the (precise) sum/value? $\endgroup$ Nov 13, 2020 at 8:58

1 Answer 1


At least, you could show the convergence. Consider the more general case of $$a_n=\frac{k^{n\,\log(n)}}{n!}\implies \log(a_n)=n\log(n)\log(k)-\log(n!)$$ Using Stirling approximation $$\log(a_n)=n ((\log (k)-1) \log (n)+1)-\frac{1}{2} \log (2 \pi n)-\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ Use Taylor approximation to get $$\log(a_{n+1})-\log(a_n)=((\log (k)-1) \log (n)+\log (k))+\frac{\log (k)-2}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$\frac {a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n) }=k\,n^{\log (k)-1}\left(1+\frac{\log (k)-2}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$ which tends to $0$ if $k<e$ which is the case for $k=2$.

Now, we know that $\sum_{n=2}^\infty a_n$ will converge to ... a number but, as already said in comments, I am quite skeptical about a possible closed form.

As also said in comments, the partial sums $$S_p=\sum_{n=2}^p \frac{2^{n\,\log(n)}}{n!}$$ converge quite fast $$\left( \begin{array}{cc} p & S_p \\ 10 & 19.337809138017622646 \\ 100 & 33.087403209274342050 \end{array} \right)$$ This value with $20$ significant figures is obtained for $k=97$.

Inverse symbolic calculators do not recognize it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .