<s>[INST] Find $N \in \mathbb{N}$ so for $n \geq N$: $ \frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}. $ Show that there exists some $N \in \mathbb{N}$ so that for all $n \geq N$:
$$
\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{4}{25}\frac{\log(n)^2}{n^{1/5}}.
$$
Numerically I know that the statement is true for $N = 6$ but if you can show it for some $N \geq 6$, that would be helpful too.
I used that $\frac{n}{n-1} \leq \frac{6}{5}$ for all $n \geq 6$ and that $\sqrt{2^n} \geq n^{1/5}$ to get
$$
\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \leq \frac{6}{5}\frac{\log (n)}{n^{1/5}} 
$$
but I don't know how to proceed from here. I think maybe I've been too rough in my estimation.
Any ideas?
 [/INST] We have
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}
\sim\frac{25}4\frac{n^{1/5}}{\log(n)}2^{-n/2}\xrightarrow{n\to\infty}0$$
hence, by definition of limit, there exists $N\in\Bbb N$ such that
$$\frac{n}{n-1} \frac{\log(n)}{\sqrt{2^n}} \frac{25}4\frac{n^{1/5}}{\log(n)^2}\leq 1$$
for every $n\geq N$.
To prove that we can choose $N=6$ consider for $x\geq 6$:
\begin{align}
0\leq\frac{x}{x-1} \frac{\log(x)}{\sqrt{2^x}} \frac{25}4\frac{x^{1/5}}{\log(x)^2}
&\leq \frac{6}{5} \frac{\log(x)}{\sqrt{2^x}} \frac{25}4\frac{x^{1/5}}{\log(x)^2}\\
&=\frac{15}{2}\frac{1}{\sqrt{2^x}}\frac{x^{1/5}}{\log(x)}\\
&=f(x)
\end{align}
do that 
\begin{align}
\frac{f'(x)}{f(x)}
&=\frac 1{5x}-\frac 1{x\log(x)}-\frac 12\log(2)\\
&\leq\frac 1{30}-\frac 12\log(2)<0
\end{align}
so that $f'(x)<0$ for $x\geq 6$, hence $f(x)\leq f(6)\leq 1$ for $x\geq 6$.
</s>