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

Question

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

Answer 1

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.

Answer 2

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 $(4)$.

Answer 3

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

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

Answer 4

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

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

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

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

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

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

Answer 5

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