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Could please give me a hint.

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

It looks like an integral, but I failed to figure out the function.

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Let $a_k = \dfrac{k}{k^2+1} \Rightarrow L = \displaystyle \lim_{n\to \infty} \dfrac{a_1+a_2+\cdots + a_n}{n} = \displaystyle \lim_{n\to \infty} a_n = 0$, by Cesaro theorem.

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HINT: For any $k\in \mathbb{N}$: $$ \frac{k}{k^2+1}\leq \frac{k}{k^2}\leq \frac{1}{k}\leq \int_{k-1}^k\frac{1}{t}\, dt. $$

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If you want it in an integral form, here's your limit.

$L=\lim_{n \to \infty} \frac{1}{2n}\int^1_0 \frac{2x \;dx}{x^2+\frac{1}{n^2}}=\lim_{n \to \infty} \frac{1}{2n} \left|\ln \left|x^2+\frac{1}{n^2}\right|\right|_0^1$

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  • $\begingroup$ Where does the $\int_0^1$ come from? Are you saying this $L$ is the same value of the sum in OPs question? $\endgroup$ – Nikolaj-K Jul 16 '15 at 9:25

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