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What is the closed form sum of this series?

$(1 - \frac12)+(\frac13 - \frac14)(1 - \frac12 + \frac13)+(\frac15 - \frac16)(1 - \frac12 + \frac13 - \frac14 + \frac15)+(\frac17 - \frac18)(1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17)+...$

I have been working on this infinite series (and other similar series). Wolfram doesn't know and I don't have any other mathematical software so I worked out an answer but I'm not sure if it is correct. Therefore I would like to see what answer anyone else can get to compare.

Also I would like to see some alternative proofs even if my answer is right because my proof was geometric and involved a lot of drawing!

user1176409
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KingChem
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1 Answers1

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It looks like it's given by $$ \frac{1}{2}\left[\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots\right)^2 - \left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots\right)\right]+\left(1+\frac{1}{9}+\frac{1}{25}+\ldots\right). $$ That is, it contains just the off-diagonal terms from the square of the series for $\ln(1+x)$ (evaluated at $x=1$), plus the odd diagonal terms. This evaluates to $$ \frac{1}{2}\left[\left(\ln 2\right)^2-\frac{\pi^2}{6}\right]+\frac{\pi^2}{8}=\frac{1}{2}\left(\ln 2\right)^2+\frac{\pi^2}{24}, $$ in agreement with your result.

mjqxxxx
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  • Thanks. Is this a general way to sum series of this type? – KingChem Dec 28 '12 at 07:15
  • I don't exactly know what you mean by "series of this type"... if there's a specific family you're working on, you could make a question about it. – mjqxxxx Dec 28 '12 at 07:20
  • I'm sorry to be so vague, the problem is my lack of knowledge makes me unable to say what kind of series I'm talking about specifically. I'll give it a try. The type of series that have term coefficients that are partial sums? – KingChem Dec 28 '12 at 07:37