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
$$\frac{2^n\cdot n^5}{n!} = \frac{6^n}{n!} \cdot \left(\frac{2}{3}\right)^n<\frac{6^n}{n!} $$
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.]
I don't know "you have to use integral test " or ...? but I suggest ratio test $$\left|\frac{a_{n+1}}{a_n}\right|= \\\dfrac{\dfrac{2^{n+1}\cdot (n+1)^5}{(n+1)!}}{\dfrac{2^n\cdot n^5}{n!}}\\=\frac{2(n+1)^5}{(n+1)n^5}\\=\frac{2(n+1)^4}{n^5}\to 0$$
All solutions given are very good and effective, as a further alternative by root test we obtain
$$\sqrt[n]{\frac{2^n\cdot n^5}{n!}}=\frac{2\cdot \sqrt[n]{n^5}}{\sqrt[n]{n!}}\to 0$$
indeed by standard limits
If you really want to use the integral test, you need to find an integrable upper bound like say: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $
$\dfrac{2^n·n^5}{n!} ≤ \dfrac{2^n·n^5}{1·2^{n-8}·(n/2)^7}$ for every natural $n ≥ 4$.
Can you finish now?
