How Find this sum $$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$

where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$

My idea: since $$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$ so $$I=\sum_{n=1}^{\infty}H^3_{n}\int_{0}^{-1}x^ndx$$ then I can't.Thank you

This problem is not Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

  • $\begingroup$ I don't really know (haven't tried anything), but this looks like it could possibly be of some use... $\endgroup$ – apnorton Feb 3 '14 at 4:04
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Feb 3 '14 at 4:10
  • $\begingroup$ My guess is that it converges to $0$ -- perhaps consider the two separate series of positive and negative terms and somehow show they are equal? $\endgroup$ – MCT Feb 3 '14 at 4:25
  • $\begingroup$ @MhenniBenghorbal,But My problem is different you link problem.But Thank you all the same $\endgroup$ – math110 Feb 3 '14 at 17:14
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    $\begingroup$ I can give you the final result: $$-\frac{9}{8} \zeta (3) \log (2)+\frac{\pi ^4}{288}-\frac{\log ^4(2)}{4}+\frac{1}{8} \pi ^2 \log ^2(2)$$ Currently, I do not have the time to post a proof. $\endgroup$ – Shobhit Bhatnagar Feb 12 '14 at 12:59

We will use the combinatorial identity, which can be proved through induction

$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$

Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n$ and summing both sides from $n =0$ to $\infty$, we get

$$\sum_{n=0}^\infty\left(H_n^{(1)}\right)^3 x^{n} - 3\sum_{n=0}^\infty H^{(1)}_{n-1}H^{(2)}_{n-1} x^{n} + 2\sum_{n=0}^\infty H^{(3)}_{n-1} x^{n} = 6\sum_{n=0}^\infty\left [ n+1 \atop 4\right]\frac{x^n}{(n-1)!} \tag{1}$$

Then note that we have the generating function

$$\sum_{n=1}^\infty (-1)^{n-k}\left [ n \atop k \right] \frac{z^n}{n!} = \frac{\log(1+z)^k}{k!}$$

Assuming $k = 4$, making the sub $z \mapsto -z$ gives

$$\sum_{n=1}^\infty \left [ n \atop 4 \right] \frac{z^n}{n!} = \frac{\log(1-z)^4}{24}$$

Diffing with respect to $z$ then gives

$$\sum_{n=1}^\infty \left [ n \atop 4\right ] \frac{z^{n-1}}{(n-1)!} = -\frac{1}{6}\frac{\log(1-z)^3}{1-z} \\ \!\!\!\!\!\!\!\!\implies \sum_{n=0}^\infty \left [ n + 1 \atop 4\right ] \frac{z^n}{(n-1)!} = -\frac{1}{6}\frac{\log(1-z)^3}{1-z} \tag{2}$$

Then subbing $(2)$ to the left side of $(1)$ gives us

$$\sum_{n=0}^{\infty}\left(H_n^{(1)}\right)^3x^n = \frac{\log^3(1-z)}{1-z} + 3\sum_{n=0}^\infty H^{(1)}_{n}H^{(2)}_{n} x^n - 2\sum_{n=0}^\infty H^{(3)}_{n} x^n \tag{3}$$

The rightmost sum is simply ${\text{Li}_3(x)}/(1-x)$, by summation interchange. The middle one is tricky.

$$\begin{align} \sum_{n=1}^\infty H_{n}H_{n}^{(2)} x^n &= -\sum_{n=1}^\infty x^n H_n \left( \psi_1(n+1)-\psi_1(1) \right) \\ &=-\frac{\psi_1(1)\log(1-x)}{1-x}-\sum_{n=1}^\infty x^n H_n \psi_1(n+1) \\ &= -\frac{\psi_1(1)\log(1-x)}{1-x}+\sum_{n=1}^\infty x^n H_n \int_0^1 \frac{z^n \log(z)}{1-z}dz \\ &= -\frac{\psi_1(1)\log(1-x)}{1-x}-\int_0^1 \frac{\log(z)\log(1-zx)}{(1-z)(1-xz)}dz \end{align}$$

Which is, through partial factorization, in turn

$$\!\!\!\!\!\!\!\!\!\!-\frac{\psi_1(1)\log(1-x)}{1-x}-\frac{1}{1-x}\int_0^1 \frac{\log(z)\log(1-zx)}{1-z}dz+\frac{x}{1-x}\int_0^1 \frac{\log(z)\log(1-zx)}{1-zx}dz \tag{4}$$

Evaluating the intermediate integral can be done, but it's quite a bit of tedious so I omit it. After some calculations, you can derive using some polylog identities that

$$\!\!\!\!\!\!\!\!\!\sum_{n=0}^\infty H^{(1)}_{n}H^{(2)}_{n} x^n = \frac{\text{Li}_3(1-x)+\text{Li}_3(x)+1/2\log^2(1-x)\log(x)-\zeta(2)\log(1-x)-\zeta(3)}{1-x} \tag{5}$$

Subbing $(5)$ and the polylog identity for the rightmost sum in $(3)$ gives

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{n=0}^{\infty}\left(H_n^{(1)}\right)^3x^n = \frac{-\pi^2/2\log(1-x)+3/2\log^{2}(1-x)\log(x)-\log^{3}(1-x)+\text{Li}_{3}(x)+3\text{Li}_{3}(1-x)-3\zeta(3)}{1-x}\tag{6}$$

Integrating with respect to $x$ and setting $x = -1$, carefully choosing the correct branch of logarithm, will give a closed form.

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  • $\begingroup$ I might have an easier approach , I'll see if it works . $\endgroup$ – Zaid Alyafeai Feb 6 '14 at 22:40
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    $\begingroup$ I wonder if there are convergence issues here, since $$\sum_{n=1}^{+\infty}H_n H_n^{(2)}x^n$$ is not a converging series in the usual sense when $|x|=1$. $\endgroup$ – Jack D'Aurizio Feb 7 '14 at 14:36
  • $\begingroup$ There could be. But there is always analytic continuation at hand. $\endgroup$ – Balarka Sen Feb 7 '14 at 14:45
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    $\begingroup$ I agree, but how can you integrate $$f(x)=\frac{\log^2(1-x)\log x}{1-x}$$ in a neighbourhood of $x=-1$ where the canonical real logarithm is not defined? In other words, what is the "correct" branch of the logarithm you claim to be taken to solve the problem? $\endgroup$ – Jack D'Aurizio Feb 7 '14 at 14:58
  • $\begingroup$ @JackD'Aurizio Well, I haven't, actually. I will be happy to see anyone finishing along that line though. I have gone through a check on Mathematica and have noted that the usual branch is not what we want here. The real part matches well though. $\endgroup$ – Balarka Sen Feb 7 '14 at 15:29

Using the identities proved in this answer, we can state: $$\frac{1}{4}\log^4(1-x)=\sum_{n=3}^{+\infty}\frac{H_n^3+2 H_n^{(3)}-3 H_n H_n^{(2)}}{n}\,x^{n+1}.\tag{1}$$ Since $$\sum_{n=1}^{+\infty}H_n^{(3)}x^n = \frac{\operatorname{Li}_3(x)}{1-x},$$ we have: $$\sum_{n=1}^{+\infty}\frac{H_n^{(3)}}{n+1}x^{n+1}=-\frac{1}{2}\operatorname{Li}_2^2(x)-\log(1-x)\operatorname{Li}_3(x),\tag{2}$$ and we only need to compute $$S=\sum_{n=1}^{+\infty}\frac{H_n H_n^{(2)}}{n}(-1)^{n+1}.$$ This is quite a difficult task. I managed to prove, through Euler's identity, that: $$f(x)=\sum_{n=1}^{+\infty}\frac{H_n^{(2)}}{n}x^n = \operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\zeta(2)\log(1-x)-\operatorname{Li}_2(1-x) \log(1-x)-2\zeta(3),\tag{3}$$ since the LHS is a primitive of $\frac{\operatorname{Li}_2(x)}{x}+\frac{\operatorname{Li}_2(x)}{1-x}$. Since $H_n=\int_{0}^{1}\frac{x^n-1}{x-1}dx$, we have: $$ S = \int_{0}^{1}\frac{f(-x)-f(-1)}{1-x}dx \tag{4}, $$ and we can probably use Landen identity in order to write $(3)$ in a nicer form and compute $(4)$.

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I'll give integral representation for Jack D'Aurizio suggestion

We have the following Nielsen formula

$$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dt=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

where we define

$$f(x)=\sum_{n\geq 0} a_n x^n$$

Hence we have

$$\int^1_0 xf(xt)\, \mathrm{Li}_2(t)\, dt=\frac{\pi^2}{6}\int^x_0 f(t)\, dt -\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

Integrating by parts we have

$$\int^1_0 F(xt)\,\frac{\log(1-t)}{t} dt+F(x)\mathrm{Li}_2(1)=\frac{\pi^2}{6}\int^x_0 f(t)\, dt -\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

Hence reducing that to

$$\int^1_0 F(xt)\,\frac{\log(1-t)}{t} dt=-\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

Differentiating w.r.t to $x$ we have

$$\int^1_0 f(xt)\,\log(1-t) \, dt=-\frac{1}{x}\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n}x^n$$

$$\int^1_0 \frac{\mathrm{Li}_2(xt)}{1-xt}\,\log(1-t) \, dt=-\frac{1}{x}\sum_{n\geq1}\frac{H_{n-1}^{(2)} H_{n}}{n}x^n$$

Let $x=-1$ to obtain

$$\sum_{n\geq1}\frac{H_{n-1}^{(2)} H_{n}}{n}(-1)^n=\int^1_0 \frac{\mathrm{Li}_2(-t)\log(1-t)}{1+t}\, \, dt$$

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  • $\begingroup$ @Jack D'Aurizio $\endgroup$ – Zaid Alyafeai Feb 8 '14 at 15:12
  • $\begingroup$ Nice(+1). ${{{{}{}}}}$ $\endgroup$ – Balarka Sen Feb 8 '14 at 15:42
  • $\begingroup$ @BalarkaSen, a standard way to evaluate non-linear sum. $\endgroup$ – Zaid Alyafeai Feb 8 '14 at 16:20
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    $\begingroup$ OMG if that's standard, then I dare not to see what's advanced! $\endgroup$ – Balarka Sen Feb 8 '14 at 16:23
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    $\begingroup$ @JackD'Aurizio , yup, multiply by $H^{(2)}_k t^k $ and sum with respect to $k$ $\endgroup$ – Zaid Alyafeai Feb 9 '14 at 16:55

We proved here that $$\small{\sum_{n=1}^\infty x^nH_n^3=\frac1{1-x}\left(\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)+\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)-3\zeta(3)\right)}$$

Replace $x$ with $-x$ then integrate from $x=0$ to $1$, we get

\begin{align} S&=\sum_{n=1}^\infty(-1)^n\frac{H_n^3}{n+1}\\ &=\frac32\int_0^1\frac{\ln (-x)\ln^2(1+x)}{1+x}\ dx-3\zeta(2)\int_0^1\frac{\ln(1+x)}{1+x}\ dx-\int_0^1\frac{\ln^3(1+x)}{1+x}\ dx\\ &\quad+\int_0^1\frac{\operatorname{Li}_3(-x)}{1+x}\ dx+3\int_0^1\frac{\operatorname{Li}_3(1+x)}{1+x}\ dx-3\zeta(3)\int_0^1\frac{1}{1+x}\ dx\\ &=\underbrace{\frac32\int_0^1\frac{\ln (-x)\ln^2(1+x)}{1+x}\ dx}_{\large I_1}-\frac32\zeta(2)\ln^22-\frac14\ln^42\\ &\quad+\underbrace{\int_0^1\frac{\operatorname{Li}_3(-x)}{1+x}\ dx}_{\large I_2}+3\operatorname{Li}_4(2)-3\zeta(4)-3\zeta(3)\ln2\tag{1} \end{align}

For $I_1$, use $\ln(-x)=\ln x+i\pi$

\begin{align} I_1&=\frac32\int_0^1\frac{\ln (x)\ln^2(1+x)}{1+x}\ dx+\frac32i\pi\int_0^1\frac{\ln^2(1+x)}{1+x}dx\\ &\overset{IBP}{=}-\frac12\int_0^1\frac{\ln^3(1+x)}{x}\ dx+i\frac{\pi}{2}\ln^32\\ &\overset{x=\frac{1-y}{y}}{=}\frac12\int_{1/2}^1\frac{\ln^3x}{x(1-x)}\ dx+i\frac{\pi}{2}\ln^32\\ &=\frac12\int_{1/2}^1\frac{\ln^3x}{x}\ dx+\frac12\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx+i\frac{\pi}{2}\ln^32\\ &=-\frac18\ln^42+\frac12\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^3x \ dx+i\frac{\pi}{2}\ln^32\\ &=-\frac18\ln^42+\frac12\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)+i\frac{\pi}{2}\ln^32\\ &=-\frac18\ln^42+\frac12\ln^42+\frac32\ln^22\operatorname{Li}_2\left(\frac12\right)+3\ln2\operatorname{Li}_3\left(\frac12\right)+3\operatorname{Li}_4\left(\frac12\right)-3\zeta(4)+i\frac{\pi}{2}\ln^32\\ &=\boxed{3\operatorname{Li}_4\left(\frac12\right)-3\zeta(4)+\frac{21}{8}\ln2\zeta(3)-\frac34\ln^22\zeta(2)+\frac18\ln^42+i\frac{\pi}{2}\ln^32\quad} \end{align}

as for the second integral: \begin{align} I_2&\overset{IBP}{=}\ln2\operatorname{Li}_3(-1)-\int_0^1\frac{\ln(1+x)\operatorname{Li}_2(-x)}{x}\ dx\\ &=-\frac34\ln2\zeta(3)+\frac12\left.\operatorname{Li}_2^2(-x)\right|_0^1=\boxed{-\frac34\ln2\zeta(3)+\frac5{16}\zeta(4)} \end{align}

To find $\operatorname{Li}_4(2) $, we use the following polylogarithmic identity: $$\operatorname{Li}_4(x)=-\frac74\zeta(4)-\frac12\ln^2(-x)\zeta(2)-\frac1{24}\ln^4(-x)-\operatorname{Li}_4(1/x)$$

set $x=2$, we get $$\boxed{\operatorname{Li}_4(2)=2\zeta(4)+\ln^22\zeta(2)-\frac1{24}\ln^42-\operatorname{Li}_4\left(\frac12\right)-i\frac{\pi}{6}\ln^32\quad}$$

Plugging the boxed results in $(1)$, we get

$$\sum_{n=1}^\infty(-1)^n\frac{H_n^3}{n+1}=\frac5{16}\zeta(4)-\frac98\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac14\ln^42\approx -0.0641$$

but the numerical value for the sum given by wolfram $\approx -0.05526$. I checked all my steps many times and did not spot any mistake, I even checked the integrals and the real value of polylorarithm and they all matched numerically but when we collect them, the final numerical value does not match. I do not know where is the mistake but I am sure that sometimes wolfram gives the wrong numerical value.

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  • $\begingroup$ I checked with Mathematica, your numerical value is correct ($-0.0641...$) $\endgroup$ – Yuriy S Jul 24 '19 at 20:28
  • $\begingroup$ Awesome. Thanks Yuriy $\endgroup$ – Ali Shather Jul 24 '19 at 20:30

$$\sum\limits_{n = 1}^\infty {\frac{{{H^3_n}}}{{n + 1}}} {\left( { - 1} \right)^{n + 1}} = \frac{1}{4}{\ln ^4}2 + \frac{9}{8}\zeta \left( 3 \right)\ln 2 - \frac{3}{4}\zeta \left( 2 \right){\ln ^2}2 - \frac{5}{{16}}\zeta \left( 4 \right).$$

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