<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] I don't know "you have to use integral test " or ...? but I suggest ratio test
$$\left|\frac{a_{n+1}}{a_n}\right|=
\\\dfrac{\dfrac{2^{n+1}\cdot (n+1)^5}{(n+1)!}}{\dfrac{2^n\cdot n^5}{n!}}\\=\frac{2(n+1)^5}{(n+1)n^5}\\=\frac{2(n+1)^4}{n^5}\to 0$$
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