How to prove $2^n\over n^2$ diverges?

Question

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.

Answer 1

$$a_{n+1}=\frac{2^{n+1}}{(n+1)^2}=2a_{n}\left(\frac{n}{n+1}\right)^2 \to \frac{a_{n+1}}{a_n}=2\left(\frac{n}{n+1}\right)^2>1, \quad \text{for}\quad n>2$$

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 $n \ge 3$.

Answer 2

HINT: $$ \frac{2^{n+1}/(n+1)^2}{2^n/n^2}=2\frac {n^2}{(n+1)^2}>1 $$

Answer 3

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 $$ \frac{2^n}{n^2}=b_n^4 $$

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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 $$ \lim_{n\to\infty}\frac{a^n}{n^k}= \lim_{n\to\infty}\left(\frac{(1+c)^n}{\sqrt{n}}\right)^{\!2k}= \infty $$

Answer 4

$$ \lim_{n\rightarrow\infty}\frac{2^n}{n^2} =\frac{\infty}{\infty}\,{\small\{\text{L'Hopital's}\}} =\frac{\log{2}}{2}\,\lim_{n\rightarrow\infty}\frac{2^n}{n} =\frac{\infty}{\infty}\,{\small\{\text{L'Hopital's}\}} =\frac{\log^2{2}}{2\cdot1}\,\lim_{n\rightarrow\infty}2^n \color{red}{\rightarrow\infty} $$
Generally: $$ \lim_{n\rightarrow\infty}\frac{a^n}{n^k} =\frac{\log{a}}{k}\,\lim_{n\rightarrow\infty}\frac{a^n}{n^{k-1}} =\frac{\log^2{a}}{k(k-1)}\,\lim_{n\rightarrow\infty}\frac{a^n}{n^{k-2}} =\,\cdots\, =\frac{\log^k{a}}{k!}\,\lim_{n\rightarrow\infty}a^n \color{red}{\rightarrow\infty} $$