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?
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.