$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}$

Question

I came across such limit $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}$$ and don't know how to find it. A factor $\frac{1}{n}$ before the sum hints that the limit should be considered as the limit of integral sums but unfortunately I can't select smth like $f(x_k)(x_{k+1} - x_k)$.

Could you please help me to solve it? I would be grateful for any hints. Thanks a lot in advance.

user · Accepted Answer · 2018-02-09T17:20:23.170

2

By Stolz-Cesaro

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}=\lim_{n \to \infty} \frac{n+1}{(n+1)^2 + 1}=0$$

edited Feb 09 '18 at 17:20

answered Feb 09 '18 at 16:32

user

154,566

1

Thank you a lot. Extremely useful theorem, wasn't aware of it – D F Feb 09 '18 at 16:35
Yes it is very useful for this kind of limits involving sum! – user Feb 09 '18 at 16:39

$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}$

1 Answers1