<s>[INST] Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that  $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.
 [/INST] Taking the $\log$ of both sides for big enough $x$
$$\begin{align*}\log \frac{x^{2x}}{(x+1)^{x+1}}&=2x\log x-(x+1)\log(x+1)\\&=x\log \frac{x^2}{x+1}-\log(x+1)\\&\geq x\log \frac{x^2}{2x}-\log(2x)\\&=\log \frac{x^{x-1}}{2^{x+1}}.\end{align*}$$
Now take $e$ to both sides to get, for large enough $x$
$$\frac{x^{x}}{(x+1)^{x+1}}\geq \frac{x^{x-1}}{2^{x+1}}\geq\frac{2^{2x-2}}{2^{x+1}}\to\infty .$$
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