# Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) – \frac{1}{2} \log^2 2$

Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$
where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$’s are known as alternating Euler sums.

Can someone provide a nice proof that
$$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) – \frac{1}{2} \log^2 2?$$

I worked for a while on this today but was unsuccessful. Summation by parts, swapping the order of summation, and approximating $H_k$ by $\log k$ were my best ideas, but I could not get any of them to work. (Perhaps someone else can?) I would like a nice proof in order to complete my answer here.

Bonus points for proving $A(1,2) = \frac{5}{8} \zeta(3)$ and $A(2,1) = \zeta(3) – \frac{1}{2}\zeta(2) \log 2$, as those are the other two alternating Euler sums needed to complete my answer.

Added: I’m going to change the accepted answer to robjohn’s $A(1,1)$ calculation as a proxy for the three answers he gave here. Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn’s approach is the one I was originally trying. I am pleased to see that it can be used to do the $A(1,1)$, $A(1,2)$, and $A(2,1)$ derivations.

## Answer

Note that
$$\dfrac{(-1)^{k-1}}k = \int_0^1 (-x)^{k-1}dx$$
and
$$\dfrac1n = \int_0^1 y^{n-1}dy$$

For the first one,
\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\
& = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \int_0^1 (-x)^{k-1}dx \int_0^1 y^{n-1} dy\\
& = \sum_{n=1}^{\infty} \int_0^1 \dfrac{(-x)^{n-1}}{1+x}dx \int_0^1 y^{n-1} dy\\
& = \int_0^1 \int_0^1\sum_{n=1}^{\infty} \dfrac{(-xy)^{n-1}}{1+x}dx dy\\
& = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dx dy\\
& = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}dy dx\\
& = \int_0^1 \dfrac{\log(1+x)}{x(1+x)} dx\\
& = \int_0^1 \dfrac{\log(1+x)}{x} dx – \int_0^1 \dfrac{\log(1+x)}{(1+x)} dx\\
& = \dfrac{\zeta(2)}2 – \dfrac{\log^2 2}2
\end{align}

$$\int_0^1 \dfrac{\log(1+x)}{x} dx = \sum_{k=0}^{\infty} \int_0^1 \dfrac{(-1)^kx^k}{k+1} dx = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^2} = \dfrac{\zeta(2)}2$$
$$\int_0^1 \dfrac{\log(1+x)}{(1+x)} dx = \left. \dfrac{\log^2(1+x)}2 \right \vert_{x=0}^{x=1} = \dfrac{\log^2 2}2$$

For the second one,

$$A(1,2) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n$$
$$\dfrac{(-1)^{k-1}}{k^2} = \int_0^1 (-x)^{k-1} dx \int_0^1 z^{k-1} dz = (-1)^{k-1} \int_0^1 \int_0^1 (xz)^{k-1} dx dz$$
\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1\int_0^1 (-1)^{k-1} (xz)^{k-1}dxdz \int_0^1 y^{n-1} dy\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+xz} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+xz)(1+xyz)} dx dy dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz(1+xz)} dx dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz – \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{1+xz} dx dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz- \int_0^1 \dfrac{\log^2(1+z)}{2z} dz\\
& = \dfrac34 \zeta(3) – \dfrac{\zeta(3)}8\\
& = \dfrac58 \zeta(3)
\end{align}

$$\int_0^1 \int_0^1 \dfrac{\log(1+xz)}{xz} dx dz = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 \dfrac{(-1)^k (xz)^k}{k+1} dx dz = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^3} = \dfrac34 \zeta(3)$$

For the third one,
$$A(2,1) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2}$$
\begin{align}
\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k} \sum_{n=1}^k \dfrac1{n^2} & = \int_0^1 \int_0^1 \int_0^1 \sum_{k=1}^{\infty} \sum_{n=1}^k (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} (-1)^{k-1} x^{k-1} (yz)^{n-1} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \dfrac{(-xyz)^{n-1}}{1+x} dx dy dz\\
& = \int_0^1 \int_0^1 \int_0^1 \dfrac1{(1+x)(1+xyz)} dx dy dz\\
& = \int_0^1 \int_0^1 \dfrac{\log(1+xy)}{(1+x)(xy)} dx dy\\
& = \zeta(3) – \dfrac{\zeta(2) \log 2}2
\end{align}

In general, if I have not made any mistake, this can be extended to $A(p,q)$.
$$A(p,q) = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$

Proceeding along similar lines, we also get that $$B(p,q) = \sum_{k=1}^{\infty} \dfrac{H_k^{(p)}}{k^q} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$

We also get that $$C(p,q) = \sum_{k=1}^{\infty} \dfrac1{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1-x_1 x_2 \cdots x_q)(1+x_1 x_2 \cdots x_{p+q})}$$
$$D(p,q) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{i=1}^k \dfrac{(-1)^{i-1}}{i^p} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q}}{(1+x_1 x_2 \cdots x_q)(1-x_1 x_2 \cdots x_{p+q})}$$

By the same argument as above, in general, nested sums like $$\sum_{k=1}^{\infty} \dfrac{(\pm 1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{(\pm 1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{(\pm 1)^{m-1}}{m^r} \cdots$$ equals
$$\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q+r+\cdots \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r+\cdots}}{(1\mp x_1 \cdots x_q)(1(\mp)(\pm)x_1 \cdots x_{p+q}) \cdots (1(\mp)(\pm)\cdots(\pm)x_1 \cdots x_{p+q+r+\cdots})}$$

For instance,
$$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{1}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1+ x_1 \cdots x_q)(1-x_1 \cdots x_{p+q}) \cdots (1-x_1 \cdots x_{p+q+r})}$$
$$\sum_{k=1}^{\infty} \dfrac{1}{k^q} \sum_{n=1}^k \dfrac{(-1)^{n-1}}{n^p} \sum_{m=1}^n \dfrac{1}{m^r} = \underbrace{\int_0^1 \cdots \int_0^1}_{p+q+r \text{ times}} \dfrac{dx_1 dx_2 \cdots dx_{p+q+r}}{(1- x_1 \cdots x_q)(1+x_1 \cdots x_{p+q}) \cdots (1+x_1 \cdots x_{p+q+r})}$$

Similarly, for negative $p$,$q$ $r$ etc, we can replace the integrals $\int_0^1$ by the appropriate differentiation operator evaluated at $1$. I will post this in detail sometime over the weekend.

Attribution
Source : Link , Question Author : Mike Spivey , Answer Author :
12 revs