4
$\begingroup$

The series arose in the calculation of Mean value of a function associated with continued fractions: $$C=\sum_{1\le b\le d<\infty}\frac{1}{b(b+d)d^2}.$$ Obviously $C=C_1-C_2,$ where $$C_1=\sum_{1\le b\le d<\infty}\frac{1}{bd^3},\quad C_2=\sum_{1\le b\le d<\infty}\frac{1}{(b+d)d^3}.$$ Mathematica gives $C_1=\frac{\pi^4}{ 72},$ and $C_2$ can be simplifyied to $$C_2=2\int_{0}^{1}\frac{\mathrm{Li}_2(t)}{t}\log(1+t)dt$$ using the following trick: $$\frac{1}{n+1}+\ldots+\frac{1}{2n}=1-\frac12+\frac13-\frac14+\ldots-\frac{1}{2n}=\int_0^1\frac{1-t^{2n}}{1+t}dt.$$

Q: Is it possible to find closed form for $C$?

$\endgroup$

1 Answer 1

6
$\begingroup$

In terms of alternating MZV's, one has (experimentally but it should be easy to prove) $$C_2=2\zeta(-3,1)+(3/4)\zeta(4)$$ where $\zeta(-3,1)=\sum_{1\le b<d}(-1)^d/(d^3b)$. It is almost certain (proved?) that $\zeta(-3,1)$ does not have a "closed" expression.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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