<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] $$y=\frac{x^{2x}}{(x+1)^{x+1}}\implies \log(y)=2x \log(x)-(x+1)\log(x+1)$$
$$\log(y)=2x \log(x)-(x+1)\left(\log(x)+\log \left(1+\frac{1}{x}\right)\right)$$
$$\log(y)=(x-1) \log (x)-(x+1)\log \left(1+\frac{1}{x}\right)$$
Now, by Taylor series
$$(x+1)\log \left(1+\frac{1}{x}\right)=(x+1)\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \Bigg]$$
$$(x+1)\log \left(1+\frac{1}{x}\right)=1+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ Putting all together
$$\log(y)=(x-1) \log (x)-1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ which already tends to $\infty$. If you want to continue, use
$y=e^{\log(x)}$
</s>