How to prove $2^n\over n^2$ diverges? Hi i was solving an exercises and now im stuck at the part where i have to show 
that this diverges. It is kind of obvious but i still would want to prove it.
 A: $$a_{n+1}=\frac{2^{n+1}}{(n+1)^2}=2a_{n}\left(\frac{n}{n+1}\right)^2 \to \frac{a_{n+1}}{a_n}=2\left(\frac{n}{n+1}\right)^2>1, \quad \text{for}\quad n>2$$
So, $a_n$ is increasing for $n>2$ and if it converges to $L$ then
$L=2L \to L=0$
what is false because $a_1=2$, $a_2=1$, $a_3=8/9$ and $a_n$ is increasing for $n \ge 3$.
A: HINT:
$$
\frac{2^{n+1}/(n+1)^2}{2^n/n^2}=2\frac {n^2}{(n+1)^2}>1
$$
A: Prove instead that
$$
b_n=\frac{(\sqrt[4]{2})^n}{\sqrt{n}}
$$
diverges. Write $\sqrt[4]{2}=1+r$, with $r>0$. Then, by Bernoulli’s inequality,
$$
(1+r)^n\ge 1+rn
$$
so
$$
b_n\ge\frac{1+rn}{\sqrt{n}}=\frac{1}{\sqrt{n}}+r\sqrt{n}
$$
and so, by comparison,
$$
\lim_{n\to\infty}b_n=\infty
$$
Now use the fact that
$$
\frac{2^n}{n^2}=b_n^4
$$

Note that this proof can be generalized to show that, for every $a>1$ and every $k>0$,
$$
\lim_{n\to\infty}\frac{a^n}{n^k}=\infty
$$
Just consider $1+c=a^{1/(2k)}$, where $c>0$, and
$$
\frac{(1+c)^n}{\sqrt{n}}\ge\frac{1}{\sqrt{n}}+c\sqrt{n}\to\infty
$$
so also
$$
\lim_{n\to\infty}\frac{a^n}{n^k}=
\lim_{n\to\infty}\left(\frac{(1+c)^n}{\sqrt{n}}\right)^{\!2k}=
\infty
$$
A: $$ \lim_{n\rightarrow\infty}\frac{2^n}{n^2} =\frac{\infty}{\infty}\,{\small\{\text{L'Hopital's}\}} =\frac{\log{2}}{2}\,\lim_{n\rightarrow\infty}\frac{2^n}{n} =\frac{\infty}{\infty}\,{\small\{\text{L'Hopital's}\}} =\frac{\log^2{2}}{2\cdot1}\,\lim_{n\rightarrow\infty}2^n \color{red}{\rightarrow\infty} $$ 
   
Generally: 
$$ \lim_{n\rightarrow\infty}\frac{a^n}{n^k} =\frac{\log{a}}{k}\,\lim_{n\rightarrow\infty}\frac{a^n}{n^{k-1}} =\frac{\log^2{a}}{k(k-1)}\,\lim_{n\rightarrow\infty}\frac{a^n}{n^{k-2}} =\,\cdots\, =\frac{\log^k{a}}{k!}\,\lim_{n\rightarrow\infty}a^n \color{red}{\rightarrow\infty} $$ 
