Alternative way to solve a limit problem $$
\lim _{n \rightarrow \infty} \frac{1}{1+n^{2}}+\frac{2}{2+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
I want to find the limit of this infinite series which I found in a book. The answer is $1/2$.
The solution to this limit was given by Sandwich/Squeeze Theorem, which was basically that the above function lies between:
$$
\frac{1}{n+n^{2}}+\frac{2}{n+n^{2}}+\frac{3}{n+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
And,
$$
\frac{1}{1+n^{2}}+\frac{2}{1+n^{2}}+\cdots+\frac{n}{1+n^{2}}
$$
series and the limit of both of these series tend to $1/2$ as $n \to \infty$.
I fully understood the solution, but I find that this isn't something that naturally/intuitively comes to your mind. I mean we need to find two different series by trial and error, both of which need to converge to a single number.
Is there any different solution to this limit problem, like dividing by powers of n, or maybe telescoping sums?

This answer uses the same idea as the answers of TurlocTheRed and Koro, but shows that one doesn't need infinite series or big-$O$ notation. Each denominator is extremely close to $n^2$, so let's compare explicitly:
\begin{align*}
\sum_{k=1}^n \frac{k}{k+n^2} &= \sum_{k=1}^n \frac{k}{n^2} - \sum_{k=1}^n \bigg( \frac{k}{n^2} - \frac{k}{k+n^2} \bigg) \\
&= \frac1{n^2} \sum_{k=1}^n k - \sum_{k=1}^n \frac{k^2}{n^2(k+n^2)}.
\end{align*}
By standard formulas, the first sum is exactly $\frac12(1+\frac1n)$; while the second sum is positive and bounded above by the sum
$$
0 < \sum_{k=1}^n \frac{k^2}{n^2(k+n^2)} < \sum_{k=1}^n \frac{k^2}{n^2(0+n^2)} = \frac1{n^4} \sum_{k=1}^n k^2 = \frac1{6n}\bigg(1+\frac1n\bigg)\bigg(2+\frac1n\bigg).
$$
The desired limit now follows from the squeeze theorem.
