I need to examine whether the following limit exists, or not. $$\lim_{n \to +\infty} \frac{1}{n^2} \sum_{k=1}^{n} k \ln\left( \frac{k^2+n^2}{n^2}\right )$$ If it does, I need to calculate its value.

How to even start this? I've got no idea.

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    $\begingroup$ This is in fact a definition of Rieman integral for $\int_{0}^{1}x \log (1+x^2) dx$ $\endgroup$ – Alex Jun 7 '16 at 15:11
  • $\begingroup$ I have deleted my answer as there is an obvious error. $\endgroup$ – Kushal Bhuyan Jun 7 '16 at 16:01
  • $\begingroup$ As an addition: $0\le \int\limits_0^1 x\ln(1+x^2) dx\le (\ln 2)\int\limits_0^1 dx=\ln 2$ $\endgroup$ – user90369 Jun 7 '16 at 17:37

An idea: "Riemann sums" may be a good start.

Massage your current sum into something of the form $$ \frac{1}{n}\sum_{k=0}^n \frac{k}{n} \ln \left( 1+\left(\frac{k}{n}\right)^2\right) $$ and recognize a Riemann sum for the (continuous) function $f\colon[0,1]\to\mathbb{R}$ defined by $f(x) = x\ln(1+x^2)$.

Update: Jack d'Aurizio gave a way (actually, two) to evaluate the integral $$\int_0^1 x\ln(1+x^2)dx$$ in his separate answer, which complements this one.

  • $\begingroup$ As for computing this integral $\int_0^1 f$... I would probably first try to expand $f$ as a power series (which is straightforward) and then argue about termwise integration to get something like $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k(k+1)}$, which is a sum supposedly quite easier to handle. But there may be much simpler: marking this answer [Community Wiki] if someone wants to give it an (elegant) shot. $\endgroup$ – Clement C. Jun 7 '16 at 14:28
  • $\begingroup$ I computed this integral for x and I got $\dfrac{\left(x^2+1\right)\ln\left(x^2+1\right)-x^2}{2}+C$, so the definite integral will be $\dfrac{2\ln\left(2\right)-1}{2}$ - a bit different that one caluculated using Stolz-Cesaro... $\endgroup$ – piternet Jun 7 '16 at 14:50
  • $\begingroup$ I have to check (currently from a cellphone), but I am no longer clear on why the difference of the a_n has this nice form for Stoltz-Cesaro. The n inside the logarithm is not the same, it should become n+1. @piternet $\endgroup$ – Clement C. Jun 7 '16 at 14:59
  • $\begingroup$ I have deleted my answer as there is an obvious error. $\endgroup$ – Kushal Bhuyan Jun 7 '16 at 16:01

Just to complete Clement C.'s answer, integration by parts leads to:

$$\begin{eqnarray*} I=\int_{0}^{1}x \log(1+x^2)\,dx &=& \left.\frac{x^2}{2}\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{x^3}{1+x^2}\,dx\\&=&\frac{\log(2)}{2}-\int_{0}^{1}x\,dx+\int_{0}^{1}\frac{x\,dx}{1+x^2} \\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$

Another chance is given by termwise integration of a Taylor series: $$\begin{eqnarray*}I = \int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}x^{2n+1}}{n}\,dx&=&\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n(n+1)}\\&=&\frac{1}{2}\left(\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}-\sum_{n\geq 1}\frac{(-1)^{n+1}}{n+1}\right)\\&=&\frac{1}{2}\left(2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}-1\right)=\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$

A third way is given by the substitution $x=\sqrt{z-1}$ plus Feynman's trick, from which: $$\begin{eqnarray*} I = \frac{1}{2}\int_{1}^{2}\log(z)\,dz = \left.\frac{1}{2}\frac{d}{d\alpha}\int_{1}^{2}z^{\alpha}\,dz\,\right|_{\alpha=0}&=&\left.\frac{1}{2}\frac{d}{d\alpha}\frac{2^{\alpha}-1}{\alpha}\,\right|_{\alpha=1}\\&=&\left.\frac{1+2^a(a\log 2-1)}{2a^2}\right|_{\alpha=1}\\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$

  • $\begingroup$ Just for completeness (this is the point I had left out in the comment after my answer): for the termwise integration, there is a little bit more justification needed, isn't there?Namely, the radius of convergence of the power series is $1$, so either doing the whole thing on $[0,r]$ and then taking the limit as $r\to 1^-$, or arguing somehow uniform convergence on $[0,1]$ (there is no normal convergence, but is there uniform convergence?) seems required. $\endgroup$ – Clement C. Jun 7 '16 at 21:05
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    $\begingroup$ @ClementC.: you are right, there is a hidden application of Abel's lemma (summation by parts) in the second approach, that is just the "discrete counterpart" of the first approach (in which we use integration by parts, of course). $\endgroup$ – Jack D'Aurizio Jun 7 '16 at 21:07
  • $\begingroup$ Thank you. (Linked your answer from mine, as complementary). $\endgroup$ – Clement C. Jun 7 '16 at 21:09
  • $\begingroup$ @ClementC.: nice idea, I did the same. $\endgroup$ – Jack D'Aurizio Jun 7 '16 at 21:10

Another approach. Using Abel's summation we have $$S=\sum_{k=0}^{n}k\log\left(1+\left(\frac{k}{n}\right)^{2}\right)=\frac{n\left(n+1\right)\log\left(2\right)}{2}-\int_{0}^{n}\frac{\left\lfloor t\left(t+1\right)\right\rfloor t}{n^{2}+t^{2}}dt $$ where $\left\lfloor x\right\rfloor $ is the floor function. Since $\left\lfloor x\right\rfloor =x+O\left(1\right) $ we have $$S=\frac{n\left(n+1\right)\log\left(2\right)}{2}-\int_{0}^{n}\frac{t^{2}\left(t+1\right)}{n^{2}+t^{2}}dt+O\left(1\right) $$ and the integral is not too complicated $$\int_{0}^{n}\frac{t^{2}\left(t+1\right)}{n^{2}+t^{2}}dt=-n^{2}\int_{0}^{n}\frac{t}{n^{2}+t^{2}}dt-n\int_{0}^{n}\frac{1}{n^{2}+t^{2}}dt+\int_{0}^{n}tdt+\int_{0}^{n}1dt $$ $$=-\frac{1}{4}n^{2}\log\left(4\right)+\frac{n^{2}}{2}-\frac{\pi}{4}n+n $$ hence $$\frac{1}{n^{2}}\sum_{k=0}^{n}k\log\left(1+\left(\frac{k}{n}\right)^{2}\right)=\log\left(2\right)+\frac{\log\left(2\right)}{n}-\frac{1}{2}-\frac{\pi}{4n}+\frac{1}{n}+O\left(\frac{1}{n}\right)\rightarrow\log\left(2\right)-\frac{1}{2} $$ as $n\rightarrow\infty$.


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