Hi there,

I was wondering if you guys could be able to find the sum of the following series:

$ S = 1/((1\cdot2)^2) + 1/((3\cdot4)^2) + 1/((5\cdot6)^2) + ... + 1/(((2n-1)\cdot2n)^2) $, in which $\{n\to\infty}$ .

This question came to mind when I was looking at this (http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf) paper by Professor Anirban DasGupta. In the last section, a couple of specific examples of his 'unified' method to find the sums of infinite series is pressented. In equation (34), he states that the following series:

$ 1/(1\cdot2) + 1/(3\cdot4) + 1/(5\cdot6) + ... 1/(2n\cdot(2n-1)) = log(2) $ (Note that $\{n\to\infty}$ again). I was wondering If it's possible to find the sum if the values of the denominators of the terms are squared.

Thanks in advance,

Max Muller

  • $\begingroup$ The second sum which you have written does not sum to $\log 2$. It is an alternating series which sums to 1. $\endgroup$ – Peter Luthy May 11 '10 at 16:59
  • $\begingroup$ Maple gives the value $-3+\pi^2/3$. More generally, let $f(r) = 1/2^r + 1/12^r + 1/30^r + \cdots$, so your sum is $f(2)$. Then Maple gives $f(3) = 10-\pi^2, f(4) = -35+10\pi^2/3+\pi^4/45$, and in general $f(r)$ is a linear combination with rational coefficients of even powers of $\pi$. There appears to be some general theory at work here but I don't know what it is, which is why this is a comment and not an answer. $\endgroup$ – Michael Lugo May 11 '10 at 16:59
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    $\begingroup$ Michael, your "general theory" is simply that 1/(x(x+1))^n is a sum of a bunch of x^j and (x+1)^j, for j varying over even integers between -n and -1. Just subtract off the principal parts of the Laurent series around 0 and -1; that gives an entire function which decays at infinity, hence equals zero. $\endgroup$ – David Hansen May 11 '10 at 17:09
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    $\begingroup$ That being said, my original comment is in error; I was summing every term, and we only want every other term. $\endgroup$ – Michael Lugo May 11 '10 at 17:39
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    $\begingroup$ Ah, I see, you have written the general term down incorrectly, Max. It is not $1/(n(n+1))^2$ . It should be $1/(2n(2n-1))^2$. $\endgroup$ – Peter Luthy May 11 '10 at 17:49

Yes, it equals $\frac{\pi^2}{3}-3$. This follows from applying partial fractions and using $\zeta(2)=\frac{\pi^2}{6}$.

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    $\begingroup$ haha I had just written this up a second before the site indicated you posted your solution... here is what I wrote: Write $\frac{1}{n^2(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\frac{1}{n^2}-\frac{2}{n(n+1)}+\frac{1}{(n+1)^2}$. Summing the first term gives $\pi^2/6$, the last term gives $-1+\pi^2/6$, and the middle term is an alternating seriesgives $-2$, so that the whole sum is $\frac{\pi^2}{3}-3$. $\endgroup$ – Peter Luthy May 11 '10 at 16:57
  • $\begingroup$ Presumably this generalizes to give the $f(r)$ I alluded to. $\endgroup$ – Michael Lugo May 11 '10 at 17:05
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    $\begingroup$ I think he's adding up over every other $n$ so it's ${\pi^2 \over 6} - 2 \log 2$. $\endgroup$ – Michael Greenblatt May 11 '10 at 17:25
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    $\begingroup$ Peter is in error, log 2 is correct. $\endgroup$ – Gerald Edgar May 11 '10 at 17:53
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    $\begingroup$ Gerald, Max wrote the incorrect general term, hence my error. $\displaystyle{\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}=\sum_{n=1}^\infty\frac{1}{(2n-1)^2}-2\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}+\sum_{n=1}^\infty\frac{1}{(2n)^2}}$. The first and second sums add to give $\displaystyle{\sum_{n=1}^\infty\frac{1}{n^2}}$, which then gives Michael Greenblatt's result. $\endgroup$ – Peter Luthy May 11 '10 at 17:56

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