$$\lim_{n\to \infty } \frac 1n \sum_{k=1}^{n} k \ln\left(\frac{n^2+(k-1)^2}{n^2+k^2}\right)$$
Basically, First thought I had was to convert the limit into definite integral but couldn't do it.I don't have any other idea to solve it.
Thanks for any help in advanced.
A little help with a solution, try to complete the solution yourself, next time please write your attempt.
Hint : $$\sum_{k=1}^{n} k \ln(\frac{n^2+(k-1)^2}{n^2+k^2})＝\sum_{k=1}^{n} k \ln(n^2＋(k－1)^2)-k\ln(n^2＋k^2)\\~\\＝\ln(n^2)-\ln(n^2＋1)＋2\ln(n^2＋1)-2\ln(n^2＋4)＋3\ln(n^2＋4)-3\ln(n^2＋9)＋...＋n\ln(n^2＋(n－1)^2)-n\ln(n^2＋n^2)\\~\\＝\ln(n^2)＋\ln(n^2＋1)＋\ln(n^2＋4)＋...＋\ln(n^2＋(n－1)^2)-n\ln(2n^2)\\~\\＝-n\ln(2n^2)＋\sum_{k=1}^{n-1} \ln(n^2＋k^2)$$
Using $$ \ln(1-x)=-x+O(x^2) $$ one has \begin{eqnarray} &&\lim_{n\to \infty } \frac 1n \sum_{k=1}^{n} k \ln\left(\frac{n^2+(k-1)^2}{n^2+k^2}\right)\\ &=&\lim_{n\to \infty } \sum_{k=1}^{n} \frac kn \ln\left(1-\frac{2k-1}{n^2+k^2}\right)\\ &=&\lim_{n\to \infty } \sum_{k=1}^{n} \frac kn\left(-\frac{2k-1}{n^2+k^2}+O\bigg(\bigg(\frac{2k-1}{n^2+k^2}\bigg)^2\bigg)\right)\\ &=&-\lim_{n\to \infty } \sum_{k=1}^{n}\frac{k(2k-1)}{n(n^2+k^2)} \end{eqnarray} which can be evaluated as $\frac\pi2-2$ by Mathematica. I don't know how to get this result by hand.
