Evaluate $\lim_{n\to \infty} \left\{\left( \sum_{r=1}^n \ln (r^2+n^2)^{\frac {r}{n^2}}\right) -\frac {\ln n}{n} -\ln n\right\}$

Question

Evaluate $$\lim_{n\to \infty} \left\{\left( \sum_{r=1}^n \ln \left[(r^2+n^2)^{\frac {r}{n^2}}\right]\right) -\frac {\ln n}{n} -\ln n\right\}$$

Now this question is seemingly much humongous than I have ever solved in the limits. Moreover so much use of logarithms and the summation is so much confusing that I am not even able to guess even the first step. I have somehow tried to create a Riemann sum if possible but couldn't proceed much. Any help would be very beneficial

Answer 1

Oh yes got that. The given question can be rewritten as

\begin{align}\lim_{n\to \infty} \left\{\left( \sum_{r=1}^n \ln \left[(r^2+n^2)^{\frac {r}{n^2}}\right]\right) -\frac {\ln n}{n} -\ln n\right\} &=\lim_{n\to \infty} \left\{\frac {1}{n^2} \left[ \left(\sum_{r=1}^n r\ln (r^2+n^2)\right) -\sum_{r=1}^n r\ln n^2\right]\right\}\\ &=\lim_{n\to \infty} \frac 1n\left( \sum_{r=1}^n \frac rn\left( \ln \left(1+\frac {r^2}{n^2}\right) \right)\right)\\ &=\int_0^1 x\ln(1+x^2)dx\\ &=\ln 2-\frac 12 \end{align}