6
$\begingroup$

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!

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ Would you please share your answer? $\endgroup$ – Jonas Meyer Dec 28 '12 at 6:21
  • $\begingroup$ @JonasMeyer (π^2)/24 + ((ln2)^2)/2 $\endgroup$ – KingChem Dec 28 '12 at 6:26
  • $\begingroup$ @KingChem Would you please share your geometric argument? $\endgroup$ – user66354 Mar 19 '13 at 4:59
12
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks. Is this a general way to sum series of this type? $\endgroup$ – KingChem Dec 28 '12 at 7:15
  • $\begingroup$ 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. $\endgroup$ – mjqxxxx Dec 28 '12 at 7:20
  • $\begingroup$ 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? $\endgroup$ – KingChem Dec 28 '12 at 7:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.