# 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,

1. Taylor expanding the root and then summing the terms
2. 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.

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.

• wow lol the problem I encountered was finding some good reason to chop of the series after the first term – DDD4C4U Feb 20 at 12:45
• wait a second tho , how did you convert the summation into $\frac{(2n+1)(8n^3 + n+1}{8n^4}$? – DDD4C4U Feb 20 at 12:46

You do not need an integral. Just squeeze:

$$\frac{2n+1}{\sqrt{n^2 +n}}\leq \sum_{j=-n}^n \left(\frac{1}{\sqrt{n^2 - j}}\right) \leq \frac{2n+1}{\sqrt{n^2 -n}}$$

• how did you come up with those terms tho? – DDD4C4U Feb 20 at 8:29
• You take the smallest and the largest term in the sum. There are $2n+1$ terms in the sum. – trancelocation Feb 20 at 8:30
• I am still bit unclear on the (2n+1) term thing – DDD4C4U Feb 20 at 8:32
• $j=-n,\ldots , -1, 0 ,1 \ldots , n$. So $2n+1$ terms. Smallest term in the sum is $\frac{1}{\sqrt{n^2+n}}$. Largest term is $\frac{1}{\sqrt{n^2-n}}$. – trancelocation Feb 20 at 8:34
• You are welcome. :-) Btw. this is a standard trick for this kind of sums. So, good to remember this trick. – trancelocation Feb 20 at 8:37