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How can we evaluate the following limit $$ \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right] $$

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  • $\begingroup$ Isn't $\ln 1+\ln 2 +\ln 3+\dots+\ln n$ converges to $n\ln n-n$, and it does not seem to difficult to evaluate from here. $\endgroup$ – S.C.B. Mar 22 '16 at 9:29
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Using the first principle, $$\begin{align} \\ & \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right] \\ & =\lim_{n\to\infty}\frac{1}{n}\left[\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{2}{n}\right)+\dots+\log\left(1+\frac{n}{n}\right)\right] \\ & =\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right) \\ & =\lim_{h\to 0} h \cdot \sum_{r=1}^{n}\log\left(1+rh\right) \\ & =\int_0^1 \log\left(1+x\right) dx \\ & =\frac{1}{\ln 10}\int_0^1 \ln\left(1+x\right) dx \\ & =\frac{1}{\ln 10}\left[x\ln\left(1+x\right)-x+\ln\left(1+x\right)\right]_0^1 \\ & =\frac{1}{\ln 10}\left[(x+1)\ln\left(1+x\right)-x\right]_0^1 \\ & =\frac{2\ln\left(2\right)-1}{\ln 10} \end{align}$$

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  • $\begingroup$ @rubik Thanks for your edit.. :-) $\endgroup$ – SchrodingersCat Mar 22 '16 at 9:45
  • $\begingroup$ You're welcome! $\endgroup$ – rubik Mar 22 '16 at 10:07
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The idea is to remark that it's a Riemann sum :

$$\frac{1}{n} \sum_{k=1}^n \ln( 1 + \frac{k}{n} ) $$

So

$$\lim \frac{1}{n} \sum_{k=1}^n \ln( 1 + \frac{k}{n} ) = \int_0^1 \ln(1+x) dx = \left[ (x+1)\ln(1+x) -x \right]_0^1 = 2\ln(2) - 1$$

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  • $\begingroup$ @Tryss Instead of integrating, you have differentiated.... Thats why your answer doesn't match mine... :-) $\endgroup$ – SchrodingersCat Mar 22 '16 at 9:35
  • $\begingroup$ I think you've made a mistake. Since $\int_0^1 \ln(1+x)dx=2\ln(2)-1$ $\endgroup$ – hamid kamali Mar 22 '16 at 9:36
  • $\begingroup$ Oups, I made a mistake in the last equality. And I interpreted your log as ln. But it seems it was a base 10 logarithm, so log(x) = ln(x)/ln(10) $\endgroup$ – Tryss Mar 22 '16 at 9:36
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    $\begingroup$ @SchrodingersCat : wow, what a splendid error ! I guess I was distracted $\endgroup$ – Tryss Mar 22 '16 at 9:37

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