<s>[INST] 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?
 [/INST] 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.
</s>