# Evaluating/Simplifying a Series involving Zeta Functions

I was trying to learn how to evaluate this expression: $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}$$ And I learned that it can be easily simplified to: $$\frac{1}{64}(\zeta(3,\frac{1}{4})-\zeta(3,\frac{3}{4}))$$ using the Hurwitz Zeta function. From here, I didn't know how to further simplify it to get an answer of: $$\frac{\pi^{3}}{32}$$ On the Mathworld Wolfram page I had seen a simplification like this: $$\zeta(s,\frac{1}{2})=2^{s}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}=2^{s}(\zeta(s)-\sum_{k=1}^{\infty}\frac{1}{(2k)^{s}})=(2^{s}-1)\zeta(s)$$ But I had no idea how they even did this, or if this is somehow the way that can help to give the simplification to get the answer above.

Edit: I have realized that this is a Dirichlet Beta Function, but now I am wondering if still the original Hurwitz Zeta Function can be simplified without the use of definitions of the Beta Function. (As in, a connection to what I have written above with that simplification)

• This is a Dirichlet beta function and you obtained the specific value $\displaystyle -1-\beta(3)=-1-\frac{\pi^{3}}{32}$ (because of $2n-1$ instead of $2n+1$) Commented Aug 5 at 21:25
• Sorry, I have edited it now to be 2n+1 as 2n-1 was a mistake. I didn't even know about this beta function, and the way which I have obtained the first simplification into a Hurwitz Zeta function was through a different method of simplification which I just did on my own. Then I have one question which is to evaluate this function, do you have to evaluate the integral representation which I have found on wikipedia? Commented Aug 5 at 21:35
• I have just realized that if you solve the integral in this case for s=3, you will not even be able to solve the integral as you will get the same series stuck inside ... So how would actually solve for one of the series. Commented Aug 5 at 21:53
• $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{2k+1}}=\frac{\pi^{2k+1}}{2^{2k+2}(2k)!}|E_{2k}|$$ encyclopediaofmath.org/wiki/Euler_numbers Commented Aug 6 at 6:15
• This is interesting because I didn't see this on the dirichlet beta function page, even though they are clearly related. Commented Aug 6 at 23:37

This is a Dirichlet beta function and your result may be obtained using this answer.

We may rewrite $$\zeta\left(3, \pm \frac {1}4\right)$$ using the polygamma function :

$$\tag{1}\psi^{(m)}(z) = (-1)^{m+1} m! \,\zeta(m+1,z)$$

You want $$\displaystyle S:=\frac{1}{64}\left(\zeta\left(3,\frac{1}{4}\right)-\zeta\left(3,\frac{3}{4}\right)\right)=\frac{1}{64}\left(-\frac 12\psi^{(2)}\left( \frac {1}4\right)+\frac 12\psi^{(2)}\left( \frac {3}4\right)\right)$$

These polygamma values are well known (from Kolbig's paper ) :

\begin{align} \tag{2}\psi^{(2n)}\left(\frac 14\right) &= -2^{2n-1}\left(\pi^{2n+1}|E_{2n}|+2(2n)!(2^{2n+1}-1)\zeta(2n+1)\right)\\ \psi^{(2)}\left(\frac 14\right)&=-2\left(\pi^{3}+2(2)!(2^{3}-1)\zeta(3)\right)\\ \tag{3}\psi^{(2n)}\left(\frac 34\right) &= 2^{2n-1}\left(\pi^{2n+1}|E_{2n}|-2(2n)!(2^{2n+1}-1)\zeta(2n+1)\right)\\ \psi^{(2)}\left(\frac 34\right)&=2\left(\pi^{3}-2(2)!(2^{3}-1)\zeta(3)\right)\\ \end{align}

Subracting the first to the second and dividing by $$64$$ gives indeed $$\displaystyle\frac {\pi^3}{32}$$.

We may get the result more directly (cf Kolbig) starting with this polygamma expression : $$\tag{4}\psi^{(n)}(z) = (-1)^{n+1}\, n! \sum_{k=0}^\infty \frac{1}{(z+k)^{n+1}}$$ using the polygamma reflection formula $$6.4.7$$ (itself derived from the $$\Gamma$$ function reflection formula $$6.1.17$$) : $$\tag{5} \psi^{(n)}\left(1-z\right)+(-1)^{n+1}\psi^{(n)}\left(z\right)=(-1)^n\pi \frac {d^n}{dz^n} \cot(\pi z)$$

So that $$\tag{6}\psi^{(2)}\left( \frac {3}4\right)-\psi^{(2)}\left( \frac {1}4\right)=\pi \left.\frac {d^2}{dz^2} \cot(\pi z)\right|_{z=1/4}$$ $$-$$ For a direct derivation use (proofs here ) : $$\tag{7} \pi \cot(\pi z)=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}=\frac 1z+\sum_{n=1}^\infty \frac 1{z-n}+\frac 1{z+n}$$ and compute the second derivative at $$\displaystyle z=\frac 14$$.

• Thank you for the brilliant answer, I didn't realize the answer would be so complicated, but I guess if you memorize only 2 necessary formulas, you could do it by hand (the polygamma-hurwitz relation formula, and the reflection formula). Commented Aug 5 at 22:29
• This is not as complicated as you may fear and you may start directly with this expression of the polygamma function $$\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}}$$ (we simply subtract the positive and negative terms of your series!). Excellent continuation anyway, Commented Aug 5 at 22:37
• Yes thank you for your explanation! Commented Aug 6 at 23:36