Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the Dirichlet beta function,

$$\beta(k) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^k}$$

(The cases k = 2 is Catalan's constant.) It seems,

$$\sum_{k=2}^\infty\Big[1-\beta(k) \Big] = \frac{1}{4}\big(\pi+\log(4)-4\big)=0.131971\dots$$

or, in general, for some constant p > 0,

$$\sum_{k=2}^\infty\left[1-\sum_{n=0}^\infty\frac{(-1)^n}{(pn+1)^k} \right] = \sum_{m=1}^\infty\frac{1}{2p^2m^2+3pm+1}$$

Anyone knows how to prove the general proposed equality? (This is similar to the question on the zeta sum here.)

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Here is a way to derive a slightly different looking result:

Notice that $$\sum_{k=2}^{\infty}\left[1-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(pn+1)^{k}}\right]=\sum_{k=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{k}}$$

$$=\sum_{n=1}^{\infty}(-1)^{n-1}\sum_{k=2}^{\infty}\frac{1}{(pn+1)^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(pn+1)^{2}}\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}.$$ Now, since $$\sum_{k=0}^{\infty}\frac{1}{(pn+1)^{k}}=\frac{1}{1-\frac{1}{pn+1}}=\frac{pn+1}{pn},$$ our series is

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}.$$

Plugging in the case $p=2$ seems to agree with your first identity.

Remark: Using partial fractions, we can go a bit further. Notice that $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn(pn+1)}=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{pn}-\frac{1}{pn+1}\right)=\frac{\log 2}{p}-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{pn+1} $$

Suppose $p$ is an integer, and let $\zeta_{p}$ be a $p^{th}$ root of unity. Then consider $$\frac{\log\left(1+z\right)}{z}+\frac{\log\left(1+\zeta_{p}z\right)}{\zeta_{p}z}+\cdots+\frac{\log\left(1+\zeta_{p}^{p-1}z\right)}{\zeta_{p}^{p-1}z}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}z^{n-1}\sum_{k=0}^{p-1}\zeta_{p}^{k(n-1)} $$

$$=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}z^{pn}.$$ Letting $z=1,$ we have the identity $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{pn+1}=\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z},$$ so our original series is $$\frac{1}{p}\log2+\sum_{k=0}^{p-1}\frac{\log\left(1+\zeta_{p}^{k}\right)}{\zeta_{p}^{k}z}.$$

share|improve this answer
1  
That was fast. :-) WolframAlpha says my non-alternating series and your alternating one are in fact equal. –  Tito Piezas III Jun 24 '12 at 16:44
    
@Tito: Ahhh, good point. –  Eric Naslund Jun 24 '12 at 17:11
    
At the end, you said, "Letting $z=1$," but there's still a $z$ there. Is that a typo? –  columbus8myhw Aug 29 at 12:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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