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 $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$
 A: 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 $(4)$.
A: 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 

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

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