Evaluate $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2}$ Considering the sum as a Riemann sum, evaluate $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k}{n^2+k^2} .$$
 A: 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
A: Another approach, more quantitative than recognizing the limit as a Riemann sum, is to estimate using integral approximations. 
First note that
$$f(x)=\frac{x}{n^2+x^2}$$
has derivative
$$f'(x)=\frac{n^2-x^2}{(n^2+x^2)^2}$$
so $f$ is increasing when $x\in[0,n]$. Thus, we can estimate $\sum f(k)$ by the same kind of approach that is used in the integral test, except we deal with an increasing $f$ rather than decreasing.
For $1\leq k\leq n-1$, we have
$$\int_{k-1}^kf(x)\,dx\leq\int_{k-1}^kf(k)\,dx=f(k)=\int_k^{k+1}f(k)\,dx\leq\int_k^{k+1}f(x)\,dx$$
Summing from $k=2$ to $k=n-1$ gives
$$\int_1^{n-1}f(x)\,dx\leq\sum_{k=2}^{n-1}f(k)\leq\int_2^nf(x)\,dx$$
Adding this to the boundary cases
$$f(1)\leq f(1)\leq\int_1^2f(x)\,dx$$ 
and
$$\int_{n-1}^nf(x)\,dx\leq f(n)\leq f(n)$$
gives
$$f(1)+\int_1^{n}f(x)\,dx\leq\sum_{k=1}^{n}f(k)\leq f(n)+\int_1^nf(x)\,dx$$
But
$$\int_1^nf(x)\,dx=\frac{1}{2}\log\frac{2n^2}{n^2+1}$$
so we have the estimates
$$\frac{1}{n^2+1}+\frac{1}{2}\log\frac{2n^2}{n^2+1}\leq S_n\leq\frac{1}{2n}+\frac{1}{2}\log\frac{2n^2}{n^2+1}$$
Letting $n\to\infty$ gives $S_n\to\frac{1}{2}\log 2$.
A: $$\sum_{k=1}^n\frac{k}{n^2+k^2}=\frac{1}{n^2}\sum_{k=1}^n\frac{k}{1+\left(\frac{k}{n}\right)^2}=\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\left(\frac{k}{n}\right)^2}\xrightarrow [n\to\infty]{}\int_0^1\frac{x}{1+x^2}\,dx=\ldots$$
