# Does $\lim_{n \to +\infty} \frac{1}{n^2} \sum_{k=1}^{n} k \ln\left( \frac{k^2+n^2}{n^2}\right )$ exist?

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.

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

• 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. – Clement C. Jun 7 '16 at 14:28
• 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... – piternet Jun 7 '16 at 14:50
• 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 – Clement C. Jun 7 '16 at 14:59
• I have deleted my answer as there is an obvious error. – Kushal Bhuyan Jun 7 '16 at 16:01

$$\begin{eqnarray*} I=\int_{0}^{1}x \log(1+x^2)\,dx &=& \left.\frac{x^2}{2}\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{x^3}{1+x^2}\,dx\\&=&\frac{\log(2)}{2}-\int_{0}^{1}x\,dx+\int_{0}^{1}\frac{x\,dx}{1+x^2} \\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$
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: $$\begin{eqnarray*} I = \frac{1}{2}\int_{1}^{2}\log(z)\,dz = \left.\frac{1}{2}\frac{d}{d\alpha}\int_{1}^{2}z^{\alpha}\,dz\,\right|_{\alpha=0}&=&\left.\frac{1}{2}\frac{d}{d\alpha}\frac{2^{\alpha}-1}{\alpha}\,\right|_{\alpha=1}\\&=&\left.\frac{1+2^a(a\log 2-1)}{2a^2}\right|_{\alpha=1}\\&=&\color{red}{\log(2)-\frac{1}{2}}.\end{eqnarray*}$$
• 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. – Clement C. Jun 7 '16 at 21:05
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 $n\rightarrow\infty$.