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In question 1280454, t was asked how to find $$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}} $$ and of course, you can write this as $$ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{4 - \frac{k^2}{n^2}} = \int_0^1 \frac{dx}{4-x^2} $$ by using the variable $x=\frac{k}{n}$ in a Reimann integral.

However, $$\int_0^1 \frac{dx}{4-x^2} = \frac18\left(\ln(5)-\ln(3) \right) \approx 0.06385$$ while each of the $n$ terms in $$ \sum_{k=1}^{n} \frac{1}{4n - \frac{k^2}{n}}$$ exceeds $\frac{1}{4n}$ so the sum must be greater than $\frac{1}{4}$ in all cases. (Experimentatlly, the limit is about $0.275$)

What gives???

Mark Fischler
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  • How did you calculate the integral??? $\displaystyle \frac{1}{4-x^2}=\frac{1}{(2-x)(2+x)}=\frac{\frac{1}{4}}{2-x}+\frac{-\frac14}{2+x}$ using the Cover Up Method. – JP McCarthy May 13 '15 at 16:29
  • Note that the function $f(x)=1/(4-x^2)\gt1/4$ for all $x\in(0,1)$, so the integral $\int_0^1f(x)dx\gt1/4$ as well. – Barry Cipra May 13 '15 at 17:15

3 Answers3

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Computation error:

\begin{align} \int_0^1 \frac{dx}{4-x^2} &= \frac{1}{4}\int_0^1 \frac{1}{2+x} + \frac{1}{2-x}\,dx\\ &= \frac{1}{4} \bigl[\log (2+x) - \log (2-x)\bigr]_0^1\\ &= \frac{1}{4}\bigl[\log 3 - \log 1 - \log 2 + \log 2\bigr]\\ &= \frac{1}{4}\log 3\\ &> \frac{1}{4}. \end{align}

Daniel Fischer
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$$\dfrac1{4n-\dfrac{k^2}n}=\dfrac n{4n^2-k^2}=\dfrac14\dfrac{2n+k+(2n-k)}{(2n+k)(2n-k)}=\dfrac14\left[\dfrac1{2n+k}+\dfrac1{2n-k}\right]$$

Now follow Find the limit $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$

Community
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lab bhattacharjee
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Otherwise from your method you should take $ a$ as 2 not 4 thus the integral evaluates to be $$ \int_0^1 \frac{dx}{4-x^2} = \frac1{2a}\left(\ln(\frac{a+(x=1)}{a-(x=1)})-\ln(\frac{a+(x=0)}{a-(x=0)}) \right) = \frac1{2*2}\left(\ln(\frac{2+1}{2-1})-\ln(\frac{2}{2}) \right) = \frac1{4}\ln3 $$

Ilaya Raja S
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  • i have spent a lot of time on this question. Please do the honors. – Ilaya Raja S May 13 '15 at 17:09
  • What do you mean by "please do the honors"? – Olivier Bégassat May 13 '15 at 17:10
  • i meant a simple up vote as the answer is good and i had to browse through my text books to at last find the simple mistake Mark did. I don't mean to be rude. Not a problem i guess @OliverBegassat – Ilaya Raja S May 13 '15 at 17:13
  • I didn't take offence, but most people on here (not just you) put effort into their answers, are equally deserving of receiving upvotes yet don't ask for them. It's bad form in my opinion. – Olivier Bégassat May 13 '15 at 17:24
  • No I think you're correct to an extent but what I think is that it's how the site works you give something and you get something too – Ilaya Raja S May 13 '15 at 17:49