What is the approach to find the expression of the partial sum of a series? This question is about computing the limit of a series as $n$ goes to $\infty$.
The series $(x_n)$ is defined as:
$$\sum_{i=0}^n \frac{1}{n^2+i}$$
So the $n$th partial sum will be given by: $\frac{1}{n^2+1} + \frac{1}{n^2+2} + \frac{1}{n^2+3} + ... + \frac{1}{n^2+n}$
So $x_1 = \frac{1}{1^2 + 1}$,
$x_2 = \frac{1}{2^2 + 1} + \frac{1}{2^2 + 2}$,
$x_3 = \frac{1}{3^2 + 1} + \frac{1}{3^2 + 2} + \frac{1}{3^2 + 3}$, and so on.
I have thought about partial fraction decomposition and trying to form a telescoping series, but have been unable to re-write the expression in a way that gives such a series. I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the $n$th partial sum. If the above idea is wrong, can I have some hints as to how to approach this problem?
 A: 
I have a feeling that the limit of this series is zero, but to show it, I will need to find an expression for the  $n$ th partial sum.

Your feeling is right, but you don't need an explicit expression for the sum in order to prove it. Just note that
$$
 0 < \sum_{i=0}^n \frac{1}{n^2+i} < \sum_{i=0}^n \frac{1}{n^2} = \frac {n+1}{n^2}
$$
What happens for $n \to \infty$?
Remark: $\sum_{i=0}^n \frac{1}{n^2+i}$ is not the partial sum of an infinite series because the terms which are added depend on $n$.
A: For the limit itself, you aleady received a good answer.
We can approximate the partial sums using harmonic numbers since
$$S_n=\sum_{i=0}^n \frac{1}{n^2+i}=H_{n^2+n}-H_{n^2-1}$$ Now, 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. This could give
$$S_n=\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right)$$ Compute for $n=10$; the exact value is
$$S_{10}=\frac{61961120721796411}{590910221462160600}\approx 0.1048571$$ while the above truncated series gives
$$S_{10}\sim \frac{12583}{120000}\approx 0.1048583$$
