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.
 A: Hint:$$\frac{x^{2x}}{(x+1)^{x+1}}=x^{2x-(x+1)}\frac{1}{\left(1+ \frac{1}{x}  \right)^{x+1}} \sim\frac{x^{x-1}}{e} $$
A: 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 $x$
$$\frac{x^{x}}{(x+1)^{x+1}}\geq \frac{x^{x-1}}{2^{x+1}}\geq\frac{2^{2x-2}}{2^{x+1}}\to\infty .$$
A: $$y=\frac{x^{2x}}{(x+1)^{x+1}}\implies \log(y)=2x \log(x)-(x+1)\log(x+1)$$
$$\log(y)=2x \log(x)-(x+1)\left(\log(x)+\log \left(1+\frac{1}{x}\right)\right)$$
$$\log(y)=(x-1) \log (x)-(x+1)\log \left(1+\frac{1}{x}\right)$$
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
$y=e^{\log(x)}$
