Converting an infinite series to a definite integral The expression that I have is,
$$\lim_{n \to \infty}\sum_{j=0}^n\left(\frac{1}{\sqrt{n^2 + j}} + \frac{1}{\sqrt{n^2 - j}}\right).$$
Original expression: 
$$\lim_{n \to \infty}\sum_{j=-n}^n \left(\frac{1}{\sqrt{n^2 - j}}\right).$$
I have prove that this summation is equal to '2'
What I have tried,


*

*Taylor expanding the root and then summing the terms

*trying to convert the summation into an integral but this doesn't seem to work because of the $\frac{j}{n^2}$ term which comes when one tries to factor out the $n^2 $from the denominator. 


I had posted this problem 
Converting an infinite summation to an integral
which looks similar but doesnt not have the root expression. THis was by accident but I decided not to delete the question because people had answered already.
 A: You do not need an integral. Just squeeze:
$$\frac{2n+1}{\sqrt{n^2 +n}}\leq  \sum_{j=-n}^n \left(\frac{1}{\sqrt{n^2 - j}}\right) \leq \frac{2n+1}{\sqrt{n^2 -n}}$$
A: It is possible that I misunderstood.
Taylor expansion work quite well for this problem since, for large $n$
$$\frac{1}{\sqrt{n^2 - j}}=\frac{1}{n}+\frac{j}{2 n^3}+\frac{3 j^2}{8
   n^5}+O\left(\frac{1}{n^7}\right)$$ Keeping these terms only and summing over $j$
$$S_n=\sum_{j=-n}^n\frac{1}{\sqrt{n^2 - j}}=\frac{(2 n+1) \left(8 n^3+n+1\right)}{8 n^4}=2+\frac{1}{n}+\frac{1}{4 n^2}+\frac{3}{8 n^3}+O\left(\frac{1}{n^4}\right)$$ Computing $S_{10}\approx2.1029014$ while the expression in the middle gives $\frac{168231}{80000}=2.1028875$.
Edit
The Faulhaber's formulae used above introducing all possible powers of $n$, stopping too early the initial Taylor expansion of $\frac{1}{\sqrt{n^2 - j}}$ makes that I missed some terms. Reworking the problem and stopping at a point where the denominator should be $n^4$, the best I found is
$$S_n=\sum_{j=-n}^n\frac{1}{\sqrt{n^2 - j}}=\frac{128 n^4+64 n^3+16 n^2+24 n+15 } {64 n^4 }$$ Used for $n=10$, this would give $\frac{269171}{128000}\approx 2.1028984$ which is slightly better.
A: $$\lim_{n \to \infty}\sum_{j=0}^{n}\left(\frac{1}{\sqrt{n^{2}+j}}+\frac{1}{\sqrt{n^{2}-j}}\right)=\lim_{n \to \infty}\frac{1}{n}\sum_{j=0}^{n}\left(\frac{1}{\sqrt{1+\left(\frac{j}{n}\right)^{2}}}+\frac{1}{\sqrt{1-\left(\frac{j}{n}\right)^{2}}}\right)$$$$=\lim_{n \to \infty}\int_{0}^{1}\left(\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1-x^{2}}}\right)dx$$$$=\color{red}{2.45216991381444}$$
