<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] $$S_n=\sum_{r=1}^n \frac r{r+n^2}=\sum_{r=1}^n \frac{\frac r{n^2}}{ 1+\frac{r}{n^2}}=\sum_{r=1}^n \frac{1+\frac r{n^2}-1}{ 1+\frac{r}{n^2}}=\sum_{r=1}^n 1-\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}=n-\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}$$
$$\sum_{r=1}^n \frac{1}{ 1+\frac{r}{n^2}}=n^2\Bigg[\sum_{r=1}^{n^2+n}\frac 1r -\sum_{r=1}^{n^2}\frac 1r\Bigg]$$ So, using harmonic numbers
$$S_n=n+n^2 \left(H_{n^2}-H_{n^2+n}\right)$$ Using the asymptotics
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120
   p^4}+O\left(\frac{1}{p^{6}}\right)$$ apply it twice and continue with Taylor series to simplify.
$$H_{n^2}-H_{n^2+n}=-\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6 n^3}-\frac{1}{4 n^4}+\frac{2}{15 n^5}+O\left(\frac{1}{n^{6}}\right)$$
$$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+\frac{2}{15 n^3}+O\left(\frac{1}{n^{4}}\right)$$ which is a quite good approximation. For example,
$$S_{10}=\frac{3039003639041255}{5909102214621606}=0.514292$$ while the truncated series gives exactly $0.5143$
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