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Considering the sum as a Riemann sum, evaluate $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2} .$$

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Hint: Pull out a factor of $\frac{1}{n}$ from the summand. – Ragib Zaman Dec 12 '12 at 17:41
@RagibZaman how? – B11b Dec 12 '12 at 17:44
$\dfrac{1}{n}\dfrac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}$ – André Nicolas Dec 12 '12 at 17:47
up vote 11 down vote accepted

$$\sum_{k=1}^n\frac{k}{n^2+k^2}=\frac{1}{n^2}\sum_{k=1}^n\frac{k}{1+\left(\frac{k}{n}\right)^2}=\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}\xrightarrow [n\to\infty]{}\int_0^1\frac{x}{1+x^2}\,dx=\ldots$$

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A more general approach. When you have sums of the form $\sum_{k=0}^nf(k,n)$ and want to test convergence you may want to use the definition of Riemann Integrals. Choose your favorite partition, mine is \begin{equation}\mathcal{P}=\left\{ 0=x_0<x_1<...<\frac{i}{n}<...<x_n=1 \right\}\end{equation} Now we must choose our function $f:[0,1]\to \mathbb{R}$ wisely so that $$U_{f,\mathcal{P}}=\sum_{k=1}^n\frac{k}{n^2+k^2}$$ But \begin{equation}U_{f,\mathcal{P}}=\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}=\sum\limits_{i=1}^{n}\frac{\sup_{x\in [x_{{i-1}},x_i]}f(x)}n \end{equation} If we choose an increasing function this simplifies to $$\sum_{k=1}^n\frac{k}{n^2+k^2}=\sum_{k=1}^{n}\frac{f(x_k)}n$$ Matching the terms gives $$f(x_k)=\frac{kn}{n^2+k^2}\Rightarrow f(x)=\frac{xn^2}{n^2+n^2x^2}=\frac{x}{1+x^2}$$ That's how you can come up with $f$. The rest can be found in Antonio's answer

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