$ \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$ Evaluate:
$$ L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$$
My approach:
Each term can be written as
$$ \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{n^2+k}$$
$$ \therefore \lim_{n \to \infty}\frac k{n^2+k}=\lim_{n \to \infty}\left(1-\frac {n^2}{n^2+k}\right)=0$$ 
hence,
$$ L=0$$
Problem:
The correct answer is 1/2, please indicate the flaw in my approach or post a new solution.
Thank You
 A: HINT:
Using $\sum_{k=1}^n k=\frac{n(n+1)}{2}$ along with the estimates $n^2+1\le n^2+k\le n^2+n$ reveals
$$\frac{n(n+1)}{2(n^2+n)}\le\sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}$$
A: For a correct solution, see Mark Viola's answer. Here I explain why what you did does not yield the right answer.
The issue in your approach: each term individually goes to $0$ when $n$ goes to infinity, indeed. But you sum $n$ of them, and $n$ goes to infinity.
So you sum a large number of small terms. Essentially, you get an indeterminate form $\infty\cdot 0$, explaining the discrepancy between what you wrote and the actual answer.

Also, a quick sanity check showing the answer canno be zero: assuming for convenience $n$ is even,
You have $$\sum_{k=1}^n \frac{k}{n^2+k} \geq \sum_{k=n/2+1}^n \frac{k}{n^2+k} \geq \sum_{k=n/2+1}^n \frac{\frac{n}{2}}{n^2+n} = \frac{n}{2}\frac{\frac{n}{2}}{n^2+n} = \frac{1}{4}\frac{n^2}{n^2+n} \xrightarrow[n\to\infty]{} \frac{1}{4}
$$
so the limit, if it exists (and it does) has to be at least $1/4$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 1}^{n}{k \over n^{2} + k} & =
\underbrace{\qquad{1 \over n}\sum_{k = 1}^{n}{k \over n}\qquad}
_{\ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\large\to} \int_{0}^{1}x\,\dd x = \color{red}{1 \over 2}}}\ -\
\overbrace{\qquad\underbrace{{1 \over n^{2}}\sum_{k = 1}^{n}{k^{2} \over n^{2} + k}}_{\ds{>\ 0}}\qquad}
^{\ds{< {1 \over n^{2}}\,\pars{n^{2} \over n^{2} + 1}n}}\
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\large \to}\,\,\,
\bbx{1 \over 2}
\end{align}
