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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.

math110
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2 Answers2

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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 $w\to0$?

Nick Peterson
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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.

Claude Leibovici
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