# How to evaluate $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$.

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 $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

• I don't really know (haven't tried anything), but this looks like it could possibly be of some use... Commented Feb 3, 2014 at 4:04
• Commented Feb 3, 2014 at 4:10
• 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?
– MT_
Commented Feb 3, 2014 at 4:25
• @MhenniBenghorbal,But My problem is different you link problem.But Thank you all the same Commented Feb 3, 2014 at 17:14
• 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. Commented Feb 12, 2014 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.

• I might have an easier approach , I'll see if it works . Commented Feb 6, 2014 at 22:40
• 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 $|x|=1$. Commented Feb 7, 2014 at 14:36
• There could be. But there is always analytic continuation at hand. Commented Feb 7, 2014 at 14:45
• 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? Commented Feb 7, 2014 at 14:58
• @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. Commented Feb 7, 2014 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 $(4)$.

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

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

• @Jack D'Aurizio Commented Feb 8, 2014 at 15:12
• Nice(+1). ${{{{}{}}}}$ Commented Feb 8, 2014 at 15:42
• @BalarkaSen, a standard way to evaluate non-linear sum. Commented Feb 8, 2014 at 16:20
• OMG if that's standard, then I dare not to see what's advanced! Commented Feb 8, 2014 at 16:23
• @JackD'Aurizio , yup, multiply by $H^{(2)}_k t^k$ and sum with respect to $k$ Commented Feb 9, 2014 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. \end{align}

For the first integral, apply IBP twice using $$\int \ln(-x)/(1+x)dx=-\text{Li}_2(1+x)$$,

\begin{align} \int_0^1\frac{\ln (-x)\ln^2(1+x)}{1+x}\ dx&=-\text{Li}_2(2)\ln^2(2)+2\int_0^1 \frac{\text{Li}_2(1+x)\ln(1+x)}{1+x}dx\\ &=-\text{Li}_2(2)\ln^2(2)+2\text{Li}_3(2)\ln(2)-2\int_0^1\frac{\text{Li}_3(1+x)}{1+x}dx. \end{align}

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

Plug these integrals back,

$$S=3\ln(2)\text{Li}_3(2)-\frac32\text{Li}_2(2)\ln^2(2)+\frac{5}{16}\zeta(4)-\frac{15}{4}\ln(2)\zeta(3)-\frac32\ln^2(2)\zeta(2)-\frac14\ln^4(2).$$

Setting $$x=1/2$$ in

$$\begin{gather} \operatorname{Li}_{2a}(x)+\operatorname{Li}_{2a}\left(\frac1x\right)=\frac{i\pi\ln^{2a-1}(x)}{(2a-1)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k)!}\ln^{2k}(x),\quad x\le 1\\ \operatorname{Li}_{2a+1}(x)-\operatorname{Li}_{2a+1}\left(\frac1x\right)=\frac{i\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x).\quad x\le 1 \end{gather}$$

we have

$$\begin{gather} \operatorname{Li}_2(2)=\frac32\zeta(2)-\pi\ln(2)i\,;\\ \operatorname{Li}_3(2)=\frac{7}{8}\zeta(3)+\frac32\ln(2)\zeta(2)-\frac{\pi}{2}\ln^2(2)i\, \end{gather}$$ where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^2(2);$$ $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$$ and so

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

Addendum: A different way to show that

$$I=\int_0^1\frac{\ln(-x)\ln^2(1+x)+2\,\text{Li}_3(1+x)}{1+x}dx=\frac32\ln^2(2)\zeta(2)+\frac74\ln(2)\zeta(3)$$

without using the values of $$\text{Li}_2(2)$$ and $$\text{Li}_3(2)$$ is to consider

$$\ln(-x)\ln^2(1+x)+2\,\text{Li}_3(1+x)=\int_{-x}^1\left(2\frac{\zeta(2)-\text{Li}_2(y)}{1-y}-\frac{\ln^2(1-y)}{y}\right)dy=\int_{-x}^1 f(y)dy,$$ we have

$$I=\int_0^1 \int_{-x}^1 \frac{f(y)}{1+x}dydx.$$

Changing the order of integration as

$$\int_0^1 \int_{-x}^1 f(x,y) dydx=\int_0^1 \int_0^1 f(x,y) dxdy+\int_{-1}^0 \int_{-y}^1 f(x,y) dxdy$$

we have

$$I=I_1+I_2$$

where $$I_1=\int_0^1 f(y)\left(\int_0^1\frac{dx}{1+x}\right)dy=\ln(2)\int_0^1 f(y)dy\overset{IBP}{=}2\ln(2)\zeta(3)$$

and

$$I_2=\int_{-1}^0 f(y)\left(\int_{-y}^1\frac{dx}{1+x}\right)dy=\int_{-1}^0 f(y)\left(\ln(2)-\ln(1-y)\right)dy$$

$$=\ln(2)\int_{-1}^0 f(y)dy-\int_{-1}^0 f(y)\ln(1-y)dy$$

$$\overset{y=-x}=\ln(2)\int_0^{1} f(-x)dx-\int_0^{1} f(-x)\ln(1+x)dx$$

By integration by parts, we have

$$\int_0^{1} f(-x)dx=3\ln(2)\zeta(2)-2\int_0^1\frac{\ln^2(1+x)}{x}dx+\int_0^1\frac{\ln^2(1+x)}{x}dx$$

$$=3\ln(2)\zeta(2)-\frac14\zeta(3)$$

where we use $$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac14\zeta(3)$$

and $$\int_0^{1} f(-x)\ln(1+x)dx=\frac32\ln^2(2)\zeta(2)-\int_0^1\frac{\ln^3(1+x)}{x}dx+\int_0^1\frac{\ln^3(1+x)}{x}dx$$

$$=\frac32\ln^2(2)\zeta(2)$$

$$\Longrightarrow I_2=\frac32\ln^2(2)\zeta(2)-\frac14\ln(2)\zeta(3)$$

$$\Longrightarrow I=\frac32\ln^2(2)\zeta(2)+\frac74\ln(2)\zeta(3)$$

• I checked with Mathematica, your numerical value is correct ($-0.0641...$) Commented Jul 24, 2019 at 20:28
• Awesome. Thanks Yuriy Commented Jul 24, 2019 at 20:30

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