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I know that this limit is equal to $\frac1{2}$ but I don't know how to do it.

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Nice question (+1) – user 1618033 Jan 20 '13 at 19:57
up vote 3 down vote accepted

Recall that $\log(1-x) = -x - \dfrac{x^2}2 + \mathcal{O}(x^3)$. Hence, \begin{align} \log \left(\dfrac{n}{n+1}\right) & = \log \left(1-\dfrac1{n+1}\right)\\ & = -\dfrac1{n+1} - \dfrac1{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)^3}\right)\\ n^2\log \left(\dfrac{n}{n+1}\right) & = -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2}+ \mathcal{O} \left(\dfrac1{(n+1)}\right)\\ n + n^2\log \left(\dfrac{n}{n+1}\right) & = n -\dfrac{n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\ & = \dfrac{n^2+n-n^2}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\ & = \dfrac{n}{n+1} - \dfrac{n^2}{2(n+1)^2} + \mathcal{O} \left(\dfrac1{(n+1)}\right)\\ \end{align} Hence, the limit is $\dfrac12$.

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Shouldn't the result be $+\frac{1}{2}$? – Amzoti Jan 20 '13 at 19:53
@Amzoti Yes. Thanks corrected. – user17762 Jan 20 '13 at 19:54

Let $n=1/x $$$\lim_{n\to\infty}\left[n-{n^2}\log{\frac{n+1}{n}}\right]=\lim_{x\to0}\left[\frac{x-\log(1+x)}{x^2}\right]=\lim_{x\to0} \frac{1}{2(x+1)}=\frac{1}{2}$$ where the penultimate equality is obtained by l'Hôpital's rule.

Note: as Marvis well noticed, the first equality is true because $\lim_{x\to0}\left[\frac{x-\log(1+x)}{x^2}\right]$ exists.


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It's interesting to see your love for solving problems.Thanks! – asimath Jan 20 '13 at 20:04
Just to be picky. The fact $$\lim_{n\to\infty}\left[n-{n^2}\log{\frac{n+1}{n}}\right]=\lim_{x\to0} \dfrac{x - \log(1+x)}{x^2} (\star)$$ assumes the fact that the limit exists in the first place i.e. convergence of sequence on the left doesn't imply that the limit on the right exists, though if the limit on the right exists then it does imply that the sequence on the left converges. Hence, you need to state that since $\lim_{x\to0} \dfrac{x - \log(1+x)}{x^2}$ exists, we have $(\star)$ to be true. This is just to find fault with your nice argument :-). +1. – user17762 Jan 20 '13 at 20:16
@Chris'ssister A simple example is the sequence $\sin(2 \pi n)$. Note that $\lim_{n \to \infty} \sin(2 \pi n)$ where $n \in \mathbb{N}$ is $0$ whereas $\lim_{x \to \infty} \sin(2 \pi x)$ doesn't exist. – user17762 Jan 20 '13 at 20:38
@Marvis: You're right. This could be a problematic case. – user 1618033 Jan 20 '13 at 20:50

Since $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$ and $\log\left(\frac{n}{n+1}\right)=-\log\left(1+\frac1n\right)$,


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