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Is it possible to compute

$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^3}\ ?$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number and $H_n=\int_0^1\frac{1-x^n}{1-x}\ dx$ is the harmonic number.

The reason I wrote the harmonic number in integral representation instead of series representation is due to the non-integer argument $n/2$ of the harmonic number and as we know $H_n=\sum_{k=1}^n\frac1k$ works for only integer $n$.

A similar version $\displaystyle\small\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$ was computed here

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  • 2
    $\begingroup$ You need a similar procedure and the toughest series to compute you already find in the paper On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_ {2n}$ here researchgate.net/publication/335920055. $\endgroup$ – user97357329 Jan 22 at 8:51
  • $\begingroup$ I wish you have a different method that saves us all these tedious calculations and using all these tons of series. $\endgroup$ – Ali Shather Jan 22 at 16:41
  • $\begingroup$ If in the reduction process of the main sum, the two series in the mentioned paper appear as a sum that can be magically reduced to simpler calculations, then yes, there could be an improvement. Given the difficulty of the series, one could be very happy with having a way to go. You didn't ask for a certain type of solution, but if it is possible to calculate such a series. $\endgroup$ – user97357329 Jan 22 at 17:09
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    $\begingroup$ Here is a version similar to this one that was previously calculated in two different ways in the paper Two advanced harmonic series of weight 5 involving skew-harmonic numbers here researchgate.net/publication/337937502 $$\sum_{n=1}^{\infty} \frac{H_n \overline{H}_n}{n^3}$$ $$=\frac{1}{6}\log^3(2)\zeta (2)-\frac{7}{8}\log ^2(2)\zeta (3)+4\log(2)\zeta (4)-\frac{193 }{64}\zeta (5)-\frac{1}{60} \log ^5(2)$$ $$+\frac{3 }{8}\zeta (2) \zeta (3)+2\operatorname{Li}_5\left(\frac{1}{2}\right).$$ Both methods presented in the paper involved the calculation of advanced harmonic series. $\endgroup$ – user97357329 Jan 22 at 17:22
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    $\begingroup$ @Ali Shather Referring to your "tons of series", of which you have done an impressive lot in recent times, I would appreciate to see these classified in a general framework to get an overview what has been done and what is still open, what is easy, what is hard, which methods have proven successful. $\endgroup$ – Dr. Wolfgang Hintze Jan 23 at 8:25
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Following the same approach here

$$S=\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^3}=H_{1/2}+\sum_{n=2}^\infty\frac{\overline{H}_nH_{n/2}}{n^3},\quad H_{1/2}=2-2\ln2$$

notice that

$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n)+\sum_{n=1}^\infty f(2n+1)$$

therefore

$$S=H_{1/2}+\frac14\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{n}}{n^3}+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}H_{n+1/2}}{(2n+1)^3}$$

$$S=2-2\ln2+\frac14S_1+S_2\tag{*}$$


For $S_1$, use $\overline{H}_{2n}=H_{2n}-H_n$

$$\Longrightarrow S_1=\sum_{n=1}^\infty\frac{{H}_{2n}H_{n}}{n^3}-\sum_{n=1}^\infty\frac{H_{n}^2}{n^3}$$

$$\small{S_1=\frac{251}{16}\zeta(5)+\frac12\zeta(2)\zeta(3)+\frac83\ln^32\zeta(2)-7\ln^22\zeta(3)-\frac8{15}\ln^52-16\ln2\operatorname{Li}_4\left(\frac12\right)-16\operatorname{Li}_5\left(\frac12\right)}$$


For $S_2$, use: $$\overline{H}_{2n+1}=H_{2n+1}-H_n$$

$$H_{n+1/2}=2H_{2n}-H_n+\frac2{2n+1}-2\ln2$$

so

$$\overline{H}_{2n+1}H_{n+1/2}\\=2H_{2n}^2+H_n^2-3H_{2n}H_n-2\ln2H_{2n}+2\ln2H_n+\frac{4H_{2n}}{2n+1}-\frac{3H_n}{2n+1}-\frac{2\ln2}{2n+1}+\frac{2}{(2n+1)^2}$$

$$\Longrightarrow S_2=\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}-3\color{orange}{\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^3}}$$ $$+2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^3}}-2\ln2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}}+4\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^4}}$$ $$+2\ln2\color{blue}{\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^3}}-3\color{blue}{\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^4}}-2\ln2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^4}}_{\large \frac{15}{16}\zeta(4)-1}+2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^5}}_{\large \frac{31}{32}\zeta(5)-1}$$


The first sum is already calculated here

$$\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}=\boxed{\frac{31}{8}\zeta(5)-\frac{45}{8}\ln2\zeta(4)+\frac72\ln^22\zeta(3)-\frac78\zeta(2)\zeta(3)}$$

The orange sum is evaluated here

$$\color{orange}{\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^3}}=\boxed{\small{\frac{1}{12}\ln ^52+\frac{31}{128} \zeta (5)-\frac{1}{2} \ln ^32\zeta (2)+\frac{7}{4} \ln ^22 \zeta (3)-\frac{17}{8} \ln2\zeta (4)+2\ln2 \operatorname{Li}_4\left(\frac{1}{2}\right)}}$$

The blue sums can be calculated using the following generalization proved by @Random Variable here

$$ \sum_{n=1}^\infty\frac{H_n}{(n+a)^2}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$

so

$$\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}}=\boxed{\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)}$$

$$\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^4}}=\boxed{\frac{31}{8}\zeta(5)-\frac{15}8\ln2\zeta(4)-\frac{21}{16}\zeta(2)\zeta(3)}$$


The red ones can be evaluated using the fact that

$$2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)(1+(-1)^n)$$

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^3}}=\sum_{n=1}^\infty\frac{H_{n}^2}{(n+1)^3}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{(n+1)^3}$$

$$=\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}^2}{n^3},\quad H_{n-1}=H_n-\frac1n$$

$$=\sum_{n=1}^\infty\frac{H_n^2}{n^3}-2\sum_{n=1}^\infty\frac{H_n}{n^4}+\sum_{n=1}^\infty\frac{1}{n^5}-\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\sum_{n=1}^\infty\frac{(-1)^n}{n^5}$$

$$=\boxed{4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{155}{32}\zeta(5)+\frac{5}{8}\zeta(2)\zeta(3)+\frac74\ln^22\zeta(3)-\frac23\ln^32\zeta(2)+\frac2{15}\ln^52}$$

Similarly

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}}=\sum_{n=1}^\infty\frac{H_n}{n^3}-\sum_{n=1}^\infty\frac{1}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\sum_{n=1}^\infty\frac{(-1)^n}{n^4}$$

$$=\boxed{-2\operatorname{Li}_4\left(\frac12\right)+\frac{17}{8}\zeta(4)-\frac{7}4\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42}$$

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^4}}=\sum_{n=1}^\infty\frac{H_n}{n^4}-\sum_{n=1}^\infty\frac{1}{n^5}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\sum_{n=1}^\infty\frac{(-1)^n}{n^5}$$

$$=\boxed{\frac{93}{32}\zeta(5)-\frac32\zeta(2)\zeta(3)}$$


Combine all these results we get

$$\small{S_2=4\operatorname{Li}_5\left(\frac12\right)-\frac{713}{128}\zeta(5)+\frac{11}{16}\zeta(2)\zeta(3)+\frac{83}{16}\ln2\zeta(4)-\frac74\ln^22\zeta(3)+\frac13\ln^32\zeta(2)-\frac1{30}\ln^52+2\ln2-2}$$

Now plug $S_1$ and $S_2$ in $(*)$ we obtain that

$$\small{S=-4\ln2\operatorname{Li}4\left(\frac12\right)-\frac{211}{128}\zeta(5)+\frac{13}{16}\zeta(2)\zeta(3)+\frac{83}{16}\ln2\zeta(4)-\frac72\ln^22\zeta(3)+\ln^32\zeta(2)-\frac{5}{30}\ln^52}$$


References

$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$

$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$

$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac72\zeta(5)-\zeta(2)\zeta(3)$

$\small{\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^2}{n^3}=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{19}{32}\zeta(5)-\frac{11}8\zeta(2)\zeta(3)+\frac74\ln^22\zeta(3)-\frac23\ln^32\zeta(2)+\frac2{15}\ln^52}$

$\small{\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^32\zeta (2) -\frac{7}{8} \ln ^22\zeta (3)-\frac{1}{15} \ln ^52 -2 \ln2 \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}$

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