# Alternating Euler sums with even index

We are all aware of the generating function of $\frac{x \arctan x}{x^2+1}$ which is:

$$\frac{x \arctan x}{x^2+1} = \sum_{m=1}^{\infty} (-1)^m \left ( \mathcal{H}_{2m} - \frac{1}{2} \mathcal{H}_m \right ) x^{2m} \tag{1}$$

Now, integrating both sides, we get that:

$$\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} = \frac{\mathcal{G}}{2} - \frac{\pi \log 2}{8}$$

I do not present the calculations here of the integral , but is suffices to say that is elementary. Apply IBP and then it boils down to the integral $\int_0^{\pi/4} \log \cos x \, {\rm d}x$ which is quite known.

My question is if we can evaluate those Euler sums individually. As hard as I searched I could not find to what these two

$$\sum_{m=1}^{\infty} \mathcal{H}_{2m}x^{2m} \quad, \quad \sum_{m=1}^{\infty} \mathcal{H}_m x^{2m}$$

both converge to. I remember seeing something for the second sum , but sadly I have not kept record of the result. But never have I encountered something for the first one. I guess they are going to have closed forms.

1)

For all $|z|<1$:

$\displaystyle A(z):=\sum_{n=1}^{+\infty} H_{2n}z^{2n}$

It's well known that:

For all $|z|<1$:

$\displaystyle F(z):=\sum_{n=1}^{+\infty} H_nz^n=-\dfrac{\ln(1-z)}{1-z}$

$\displaystyle F(z)=\sum_{n=1}^{+\infty} H_{2n}z^{2n}+\sum_{n=0}^{+\infty} H_{2n+1}z^{2n+1}$

$\displaystyle G(z):=\sum_{n=0}^{+\infty} H_{2n+1}z^{2n+1}=z+\sum_{n=1}^{+\infty}\left(H_{2n}+\dfrac{1}{2n+1}\right)z^{2n+1}$

$\displaystyle G(z)=z+zA(z)+\sum_{n=1}^{+\infty}\dfrac{z^{2n+1}}{2n+1}$

$\displaystyle G(z)=zA(z)+\sum_{n=0}^{+\infty}\dfrac{z^{2n+1}}{2n+1}$

For $|z|<1$,

$\displaystyle \sum_{n=0}^{+\infty}\dfrac{z^{2n+1}}{2n+1}=\dfrac{1}{2}\ln(1+z)-\dfrac{1}{2}\ln(1-z)$

For $|z|<1$, derivative of the left member is:

$\displaystyle \sum_{n=0}^{+\infty}z^{2n}=\dfrac{1}{1-z^2}=\dfrac{1}{2}\dfrac{1}{1-z}+\dfrac{1}{2}\dfrac{1}{1+z}$

Therefore, for $|z|<1$,

$G(z)=zA(z)+\dfrac{1}{2}\ln(1+z)-\dfrac{1}{2}\ln(1-z)$

Finally, for $|z|<1$,

$A(z)=-\dfrac{1}{2}\left(\dfrac{\ln(1-z)}{1-z}+\dfrac{\ln(1+z)}{1+z}\right)$

2)

for $|z|<1$,

$\displaystyle \sum_{n=1}^{+\infty} H_nz^{2n}=F(z^2)=-\dfrac{\ln(1-z^2)}{1-z^2}$