How find this limit$\lim_{n\to\infty}\frac{1}{n}\left(\frac{n}{\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}}\right)^n$ How  find this limit
$$\lim_{n\to\infty}\dfrac{1}{n}\left(\dfrac{n}{\dfrac{1}{2}+\dfrac{2}{3}+\cdots+\dfrac{n}{n+1}}\right)^n$$
My try: since
$$\dfrac{1}{2}+\dfrac{2}{3}+\cdots+\dfrac{n}{n+1}=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)=(n+1)-H_{n+1}$$
where
$$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
then I can't.Thank you
this problem is from a book,and only give this answer
$$e^{\gamma-1}$$
where $\gamma$ is denotes the Euler–Mascheroni constant.
 A: Hint: You're on the right track!  Start by noting that if $a_n$ is the $n$-term of your sequence, then
$$
\begin{align*}
\ln a_n&=\ln\left(\frac{1}{n}\right)+n\ln\left(\frac{n}{n+1-H_{n+1}}\right)\\
&=n\ln(n)-n\ln(n+1-H_{n+1})-\ln(n)\\
&=(n-1)\ln(n)-n\ln(n+1-H_{n+1})
\end{align*}
$$
Next, note that
$$
\ln(n+1-H_{n+1})=\ln\left[n\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right)\right]=\ln n+\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right),
$$
so that
$$
\begin{align*}
\ln(a_n)&=(n-1)\ln(n)-n\ln(n)-n\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right)\\
&=-\ln(n)-n\ln\left(1+\frac{1}{n}-\frac{H_{n+1}}{n}\right).
\end{align*}
$$
Now, what do you know about asymptotics for $\frac{H_{n+1}}{n}$? How about for $\ln(1+w)$ as $w\to0$?
A: I thing that Nicholas R. Peterson gave you almost all the tricks required for your problem. I shall try to help you for the last details (since this is not homework - at least, no tagges as).
For sufficiently large values of $n$, the ratio of H(n+1)/n behaves as
(EulerGamma - Log[1/n]) / n
So, Log[1 + 1/ n - H(n+1) /n] behaves as 1  + (1 - EulerGamma + Log[1/n]) / n
So, - Log[n] - n Log[1 + 1/ n - H(n+1) /n] behaves as (-1 + EulerGamma) and, finally, $a(n)$ behaves such as Exp[EulerGamma - 1] when $n$ goes to infinite values.  
One point I would like to undeline here is that all the above has been obtained on the basis of Taylor expansions limited to first order.
