How does one easily compute the limit of $a_n=(n\cdot \ln(\frac{n+1}{n}))^n$? I can show that $\displaystyle a_n=\left(n\cdot \ln\left(\frac{n+1}{n}\right)\right)^n\rightarrow \frac{1}{\sqrt{e}}$ by expressing it as $\displaystyle e^{\ln(a_n)}$, but this ends up very tedious. What is an easier way to compute this limit?
Thanks!
Edit This is a sequence, so I mean the limit as $\displaystyle n\rightarrow \infty$.
 A: Asymptotics
$$\begin{align}
a_n &= \left(n\cdot \ln\left(\frac{n+1}{n}\right)\right)^n
= \left(n\cdot \ln\left(1+\frac{1}{n}\right)\right)^n
= \left(n\cdot \left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right)^n
\\ &= \left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)^n
= \left(\left(1-\frac{1}{n}+o\left(\frac{1}{n}\right)\right)^{1/2}\right)^{n}
= \left(\left(1-\frac{1}{n}+o\left(\frac{1}{n}\right)\right)^n\right)^{1/2}
\\ &= \left(\left(e^{-1/n}+o\left(\frac{1}{n}\right)\right)^n\right)^{1/2} = (e^{-1}+o(1))^{1/2} =e^{-1/2}+o(1) .
\end{align}$$
A: Or a little bit more explicit, based on the Taylor expansion of $\log$, the following inequalities hold for $n \geq 2$:
$$
\left(1-\frac{1}{2n}\right)^n \leq a_n \leq \left(1-\frac{1}{2n}+\frac{1}{3n^2}\right)^n \leq \left(1-\frac{1}{2(n+1)}\right)^n
$$
and both sides have limit $e^{-1/2}$.
A: You can also use the Trapezoidal rule to approximate
$$
\ln \left(\frac{n+1}{n} \right)=\int_n^{n+1}{\frac{1}{x}}dx=\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n}\right)+\mathcal{O}\left(\frac{1}{n^3}\right)
$$
Exponentiating and multiplying by $n^n$ yields
$$
n^n \ln \left(\frac{n+1}{n}\right)^n=\left[\left(\frac{2n+1}{2n+2}\right)+\mathcal{O}\left(\frac{1}{n^2}\right)\right]^n=\left(1-\frac{1}{2n}+\mathcal{o}\left(\frac{1}{n}\right)\right)^n=e^{-\frac{1}{2}}+\mathcal{o}(1)
$$
(The last equal sign is already given in the answer of GEdgar).
