Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I show that the following limit converges and $L \in (0, +\infty)$?

$\lim\limits_{n \to +\infty}\left( S_n - T_n\right)$, where $S_n = \int\limits_1^n \log x\, dx$, and $T_n = \sum_{k = 1}^{n-1}\frac{\log(k)+\log(k+1)}{2}$

I already tried using Riemann's sum, some converge tests, transforming into a telescope sum but I got nowhere. Anyone here got an idea on how to prove it?

share|improve this question

2 Answers 2

You could just evaluate it:

$$S_n=\int_1^n \ln x\;dx=n\ln n-n+1$$

$$T_n=\frac{1}{2}\sum_{k=1}^{n-1}\left(\ln k+\ln (k+1)\right)=\frac{1}{2}\left(\ln n!+\ln (n-1)!\right)=\ln \frac{n!}{\sqrt{n}}$$

Using Stirling's formula for $n!:$

$$\begin{align*}\ln \frac{n!}{\sqrt{n}}&=\ln\left[\frac{ \sqrt{2\pi n}}{\sqrt{n}}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)\right]\\&=\ln \sqrt{2\pi}+n\ln \frac{n}{e}+\ln \left(1+O\left(\frac{1}{n}\right)\right)\\&=\ln \sqrt{2\pi}+n\ln n-n+\ln \left(1+O\left(\frac{1}{n}\right)\right)\end{align*}$$


$$S_n-T_n=1-\ln \sqrt{2\pi}-\ln \left(1+O\left(\frac{1}{n}\right)\right)$$

$$\lim_{n\to\infty} S_n-T_n=1-\ln \sqrt{2\pi}$$

Since $\sqrt{2\pi}<e$ the limit is positive.

share|improve this answer

We have \begin{align}S_n&=\sum_{k=1}^{n-1}\int_k^{k+1}\log x dx=\sum_{k=1}^{n-1}\left[ x\log x -x\right]_k^{k+1}\\ &= \sum_{k=1}^{n-1}(k+1)\log(k+1)-k\log k-1, \end{align} then we find after simplification \begin{align} S_n-T_n=\sum_{k=1}^{n-1}\frac{(2k+1)\log(k+1)-(2k+1)\log k-2}{2}=\frac{1}{2}\sum_{k=1}^{n-1}u_k. \end{align} Now we prove that the serie $\sum u_n$ is convergent and positive. We have $$u_k=(2k+1)\log(1+\frac{1}{k})-2=(2k+1)(\frac{1}{k}-\frac{1}{2k^2}+\frac{1}{3k^3}+o(\frac{1}{k^3}))-2=\frac{1}{6k^2}+o(\frac{1}{k^2})$$ and by comparaison with Riemann sum we conclude.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.