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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.

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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.

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  • $\begingroup$ wow lol the problem I encountered was finding some good reason to chop of the series after the first term $\endgroup$ – DDD4C4U Feb 20 at 12:45
  • $\begingroup$ wait a second tho , how did you convert the summation into $\frac{(2n+1)(8n^3 + n+1}{8n^4} $? $\endgroup$ – DDD4C4U Feb 20 at 12:46
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You do not need an integral. Just squeeze:

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

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

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