5
$\begingroup$

Here http://integralsandseries.prophpbb.com/topic119.html

We came across the following harmonic sum

$$\tag{1} \sum_{k\geq 1}\frac{(-1)^{k-1}}{k^2}H_k^{(2)}$$

Note that we define

$$H_k^{(2)}=\sum_{n\geq 1}^k\frac{1}{n^2} $$

Also we have

$$\psi_1(k+1)= \zeta(2) -H_k^{(2)} $$

Any ideas how to evaluate (1) ?

$\endgroup$
4
$\begingroup$

A related problem. You can have the following identity

$$\sum_{k=1}^{\infty}(-1)^{k-1} \frac{H_k^{(2)}}{k^2} = \frac{37}{16}\zeta(4)+2\sum_{k=1}^{\infty}(-1)^k \frac{H_k}{k^3}\sim 0.7843781621 .$$

$\endgroup$
  • 1
    $\begingroup$ That is quite interesting formula , good job ! $\endgroup$ – Zaid Alyafeai Aug 17 '13 at 19:29
1
$\begingroup$

we know that $$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$ differentiate both sides with respect to $n$ $$\int_0^1x^{n-1}\ln(1-x)\ln x\ dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n} $$ multiply both sides by $\frac{(-1)^n}{n}$ then take the sum, we get \begin{align} \sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}+\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}-\underbrace{\zeta(2)\sum_{n=1}^\infty\frac{(-1)^n}{n^2}}_{-\frac54\zeta(4)}&=\int_0^1\frac{\ln(1-x)\ln x}{x}\sum_{n=1}^\infty\frac{(-x)^n}{n}\ dx\\ &=-\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx \end{align} I was able here to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)$$ which follows that$$\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}=-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}-\frac{11}8\zeta(4)$$ substituting the well known results: $$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$ $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li}_4\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$ finally we get$$\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)-\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42$$

$\endgroup$

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.