Show that $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \frac{1}{n}=\frac{5\zeta(3)}{8}$

Question

$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n}=\frac{5\zeta(3)}{8}$$

I tried to create a proof from some lemmas some are suggested by my Senior friends

Lemma 1 $$ {H_n} = \sum\limits_{k = 1}^\infty {\left( {\frac{1}{k} - \frac{1}{{n + k}}} \right)} = \sum\limits_{k = 1}^\infty {\frac{n}{{k \cdot \left( {n + k} \right)}}} $$

Lemma 2 $$ \begin{split} \sum\limits_{n = 1}^\infty {\frac{{{x^{2n - 1}}}}{{2n - 1}}} &= \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} - \sum\limits_{n = 1}^\infty {\frac{{{x^{2n}}}}{{2n}}} - \log \left( {1 - x} \right) - \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{x^{2n}}}}{n}} \\ &= \frac{1}{2} \cdot \log \left( {1 - {x^2}} \right) - \log \left( {1 - x} \right) \\ &= \frac{1}{2} \cdot \log \frac{{1 + x}}{{1 - x}} \end{split} $$

Lemma 3 $$ \begin{split} \int\limits_0^1 {\frac{{\log }^2 x}{1 - x}dx} & = \int\limits_0^1 {{{\log }^2}x \cdot \sum\limits_{n = 0}^\infty {{x^n}} dx} \\ &= \sum\limits_{n = 0}^\infty {\int\limits_0^1 {{{\log }^2}x \cdot {x^n} dx} } \\ &= 2 \cdot \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^3}}}} = 2 \cdot \zeta \left( 3 \right) \end{split}$$

Lemma 4 $$ \begin{split} \int\limits_0^1 {\frac{{{{\log }^2}x}}{{1 + x}} dx} & = \int\limits_0^1 {{{\log }^2}x \cdot \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n} \cdot {x^n}} dx} \\ & = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n} \cdot \int\limits_0^1 {{{\log }^2}x \cdot {x^n} dx} } \\ & = 2 \cdot \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n} \cdot }}{{{{\left( {n + 1} \right)}^3}}}} = \frac{3}{2} \cdot \zeta \left( 3 \right) \end{split}$$

Answer 1

\begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \frac{1}{n} &= \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}\, \mathrm{d}x\,\mathrm{d}z \int_0^1 y^{n-1} \, \mathrm{d}y \\ &= \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \frac{(-xyz)^{n-1}}{1+xz} \,\mathrm{d}x \, \mathrm{d}y \,\mathrm{d}z \\ &= \int_0^1 \int_0^1 \int_0^1 \frac{1}{(1+xz)(1+xyz)} \,\mathrm{d}x \,\mathrm{d}y \,\mathrm{d}z \\ &= \int_0^1 \int_0^1 \frac{\log(1+xz)}{xz(1+xz)} \,\mathrm{d}x \,\mathrm{d}z \\ &= \int_0^1 \int_0^1 \frac{\log(1+xz)}{xz} \,\mathrm{d}x \, \mathrm{d}z - \int_0^1 \int_0^1 \frac{\log(1+xz)}{1+xz} \, \mathrm{d}x \, \mathrm{d}z \\ &= \int_0^1 \int_0^1 \frac{\log(1+xz)}{xz} \,\mathrm{d}x \,\mathrm{d}z - \int_0^1 \frac{\log^2(1+z)}{2z} \, \mathrm{d}z \\ &= \frac{3}{4} \zeta(3) - \frac{\zeta(3)}{8} \\ &= \frac{5}{8} \zeta(3) \end{align}

And for the last row

$$\int_0^1 \int_0^1 \frac{\log(1+xz)}{xz} \, \mathrm{d}x \,\mathrm{d}z = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 \frac{(-1)^k (xz)^k}{k+1} \,\mathrm{d}x \,\mathrm{d}z = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^3} = \frac{3}{4} \zeta(3)$$

Answer 2

Using $$ \sum_{n=1}^k\frac1n=\int_0^1\frac{1-x^k}{1-x}dx $$ one has $$\begin{eqnarray} &&\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n}\\ &=&\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \int_0^1\frac{1-x^k}{1-x}dx\\ &=&\int_0^1\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \frac{1-x^k}{1-x}dx\\ &=&\int_0^1\frac1{1-x}\bigg(\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k^2} -\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{k^2} x^k\bigg)dx\\ &=&\int_0^1\frac1{1-x}\bigg(\text{Li}_2(-x)-\text{Li}_2(-1)\bigg)dx\\ &=&-\int_0^1\frac1{1-x}\int_{-x}^{-1}\frac{\text{Li}_1(t)}{t}dtdx\\ &=&-\int_0^1\frac1{1-x}\int_{x}^{1}\frac{\text{Li}_1(-t)}{t}dtdx\\ &=&-\int_0^1\frac{\text{Li}_1(-t)}{t}\int_{0}^{t}\frac1{1-x}dxdt\\ &=&-\int_0^1\frac{\ln(1+t)\ln(1-t)}{t}dt\\ &=&\frac58\zeta(3). \end{eqnarray}$$ Here $$\begin{eqnarray} &&\int_0^1\frac{\ln(1+t)\ln(1-t)}{t}dt\\ &=&\frac14\bigg[\int_0^1\frac{\left(\ln(1+t)+\ln(1-t)\right)^2}{t}dt-\int_0^1\frac{\left(\ln(1+t)-\ln(1-t)\right)^2}{t}dt\bigg]\\ &=&\frac14\bigg[\int_0^1\frac{\ln^2(1-t^2)}{t}dt-\int_0^1\frac{\ln^2\left(\frac{1-t}{1+t}\right)}{t}\overset{\frac{1-t}{1+t}\to t}{dt}\bigg]\\ &=&\frac14\bigg[\frac12\int_0^1\frac{\ln^2(1-t)}{t}dt-2\int_0^1\frac{\ln^2t}{1-t^2}dt\bigg]\\ &=&\cdots \end{eqnarray}$$

Answer 3

We use a radius circle N+1/2 centered at the origin of the axes as the contour.

We consider the kernel function $$\displaystyle f(z)= \frac{\pi csc(\pi z)(\gamma+\psi(-z))}{z^{2}}$$

There are poles at the positive integers , n, due to the bigamma as well as poles at the negative integers -n as well as at the origin of the axes.

Therefore the series, as $z\to n$ is $\pi csc(\pi z)(\gamma+\psi(-z))$ is

$$\displaystyle \frac{(-1)^{n}}{(z-n)^{2}}+\frac{(-1)^{n}H_{n}}{z-n}+\cdot\cdot\cdot$$

So the integral remainder is:

$$\displaystyle \lim_{z\to n}\left[Res\left(\frac{(-1)^{n}}{(z-n)^{2}}\cdot \frac{1}{n^{2}}\right)+Res\left(\frac{(-1)^{n}H_{n}}{z-n}\cdot \frac{1}{n^{2}}\right)\right]$$

So we sum the remainders and let Sthe original sum be: (Notice how we get the derivative on the left side due to a double pole)

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

$$\displaystyle = \frac{3}{2}\zeta(3)+S............. ......[1]$$

At first we get its coefficient 1/z by simply doing a Laurent expansion of it f(z)to find the residue.

$$\displaystyle \frac{1}{z^{4}}-\frac{\zeta(3)}{z}-\frac{7\pi^{4}}{360}+O(z)$$

So, $Res(f(z),0)=-\zeta(3)............[2]$

In negative integers -nthe order is, $$\displaystyle \frac{(-1)^{n}(H_{n}-1/n)}{z+n}+......$$

$$\displaystyle \lim_{z\to -n}\left[Res\left(\frac{(-1)^{n}(H_{n}-1/n)}{z+n}\cdot \frac{1}{z^{2}}\right)\right]$$

Add the residues,

$$\displaystyle S-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{3}}=S+\frac{3}{4}\zeta(3)............[3]$$

So adding [1]+[2]+[3], and solving for the original series we have:

$$\displaystyle \frac{3}{2}\zeta(3)+S+\frac{3}{4}\zeta(3)+S-\zeta(3)=0$$

$$S=\frac{-5}{8}\zeta(3)$$

Because of $(-1)^{n-1}$ in the original sum we finally get $$\displaystyle \boxed{\frac{5}{8}\zeta(3)}$$