# $\left(1+\frac{1}{n^2-1}\right)^n \le n^{\frac{1}{n}}$ for sufficiently large $n$

Is the following true $\left(1+\frac{1}{n^2-1}\right)^n \le n^{\frac{1}{n}}$ for sufficiently large $n$? I'm sure it is, but I'm having difficulty proving it. I've tried using the Bernoulli inequality and Arithmetic-Geometric mean inequality but to no avail. Is there an elegant way of doing this?

\begin{align} \left(1+\frac{1}{n^2-1}\right)^{n^2-1}&<e\\ \left(1+\frac{1}{n^2-1}\right)^{n^2}&<e\left(1+\frac{1}{n^2-1}\right)<2e<n \end{align} for $n\geq6$. Then raise both sides to the $1/n$. Then check $n=2,3,4,5$ directly. (By the way, it's false for $n=2$.)
But it's actually a pretty bad bound. This argument shows the much tighter $$\left(1+\frac{1}{n^2-1}\right)^{n}<(2e)^{1/n}$$
• The expression being in the form $(1+1/A)^B$ suggests manipulating it so that $B$ looks like $A$, and then the limit to $e$ might help. (Even if the limit were from above.) That was the motivation. Commented Nov 7, 2016 at 15:47
$n\geq 8$ grants: $$n\log\left(1+\frac{1}{n^2-1}\right) < \frac{n}{n^2-1} < \frac{2}{n} < \frac{\log(n)}{n}.$$