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

Is it the log to the $\frac{r}{n^2}$ or is it the log the sum of squares to the $\frac{r}{n^2}$? — Alex D, Apr 22 '18 at 05:49

score 2 · Accepted Answer · edited Apr 22 '18 at 06:20

2

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}

edited Apr 22 '18 at 06:20

Alex D

1,214

answered Apr 22 '18 at 06:01

Rohan Shinde

9,737

How did you get the integral? I did Laurent series expansion at $n=\infty$ and I get zero for the limit for the sum on the third line – Alex D Apr 22 '18 at 06:13
@AlexD It's just the Riemann sum. According to relation between Reimann sum and integral we have $$\lim_{n\to \infty} \frac 1n \left(f(\frac 1n)+f(\frac 2n) +\cdots +f(\frac nn) \right) =\int_0^1 f(x) dx$$. Now here $f(x)=x\ln(1+x^2)$ – Rohan Shinde Apr 22 '18 at 06:18

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

1 Answers1