<s>[INST] Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test 
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.

What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}$$
Using $$\frac{n!}{2^n}=\frac{4!}{2^4}\frac{5}{2}\frac{6}{2}\cdots \cdots >1\Longrightarrow \frac{n!}{2^n}>1$$
So $$\frac{2^n}{n!}\leq 1\Longrightarrow \frac{2^n\cdot n^5}{n!}<n^5$$
How do i solve it Help me please, Thanks
 [/INST] The integral test requires an upper bound which is not immediate. Instead we can relate the series to the Taylor expansion of the exponential.
$$n^5=n(n-1)(n-2)(n-3)(n-4)+10(n-1)(n-2)(n-3)(n-4)\\+65(n-2)(n-3)(n-4)+285(n-3)(n-4)+781(n-4)+1024$$ so that
$$\sum_{n=1}^\infty\frac{2^nn^5}{n!}=
\\\sum_{n=1}^\infty2^n\left(\frac1{(n-5)!}+\frac{10}{(n-4)!}+\frac{65}{(n-3)!}+\frac{285}{(n-2)!}+\frac{781}{(n-1)!}+\frac{1024}{n!}\right)$$ (the terms with a factorial of a negative must be omitted).
This sum is obviously convergent and equals $ae^2+b$ where $a,b$ are rational numbers.
[You can show convergence in an easier way (see @zkutch); but my goal was to show that one can find the exact value.]
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