# Handling Errors in Slowly Converging Dirichlet Beta Infinite Sum

I have tried to calculate the following slowly converging double sum in Mathematica 11.3

$$\sum _{k=1}^{\infty } \left(\frac{ 1}{(2 k-1) (2 k+1)} \left(\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{(2 n-1)^{2 k-1}} \right) \right)\tag{1}$$

where $$\beta(2k-1) =\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{(2 n-1)^{2 k-1}}$$

When using symbolic notation as in (1) Mathematica gives the incorrect symbolic answer $$\frac{3 \zeta (3)}{2 \pi ^2}$$, the correct answer being $$\frac{4}{\pi^2}\frac{ 7\, \zeta (3)}{8}\approx0.4262783988$$.

If I use the text function Sum[] instead, lots of Recursion and Iteration Depth Errors result which were presumably suppressed in the symbolic sum.

I have contacted Wolfram for clarification on this.

The problem I face in avoiding a double sum is that the standard output result for the $$\beta(k) =\sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{(2 n-1)^{k}}$$ summation in terms of the generalised Riemann Zeta Function is not valid for $$k=1$$. (The well known result for $$\beta(1)$$ is $$\beta(1)=\frac{\pi}{4}$$)

$$\beta(k)=2^{2-4 k} \left(\zeta \left(2 k-1,\frac{1}{4}\right)-\zeta \left(2 k-1,\frac{3}{4}\right)\right)\tag{2}$$

for $$k>1$$

I don't know how to force the assumptions on Sum[] to change this behaviour.

Sum[(-1)^(n - 1)/(2n-1)^k, {n, 1, Infinity}, Assumptions -> k > 0]


doesn't alter the output result. Similarly with the Dirichlet Eta Function

Sum[(-1)^(n - 1)/(n)^k, {n, 1, Infinity}, Assumptions -> k > 0]


Adding the assumption this way has no effect on the output result.

Any thoughts?

Update:

With the help of Wolfram and Somos I have clarified that the underlying problem is with the $$\beta(k)$$ summation in Mathematica, which will hopefully be fixed in future releases.

The work around is to separate the summation into two parts:

FullSimplify[Sum[(1/((2k-1)(2k+1))),{k,1,Infinity}]+Sum[(1/((2k-1)(2k+1)) Sum[((-1)^(n-1)/(2n-1)^(2k-1)),{n,2,Infinity}]),{k,1,Infinity}]]


which gives the correct symbolic result.

• If I understand your question correctly, you already know the correct answer. And Mathematica knows $\beta(1)=\pi/4$. Apr 10, 2019 at 17:45
• @Somos: It is mathematica's generalised evaluation of the $\beta(k)$ summation in terms of the generalised Riemann Zeta Function which is only valid for $k>1$. I just wondered if there was a way to enter the assumption $k>0$ somehow to change this. Apr 10, 2019 at 18:04
• @somos: Yes that's correct. The sum is fine. As soon as you convert the sum to the generalised zeta function there is a problem. The two functions don't exactly equate. Maybe I should just raise this as a separate issue with Wolfram. Apr 10, 2019 at 18:23

Using a standard definition of $$\beta()$$ by summing over odd integers we get

 Sum[(-1)^((n - 1)/2)/n^1, {n, 1, Infinity, 2}]


evalutes to $$\pi/4$$. This is a bit different than your summation, but this one works. Also, the double summation can be summed in two ways. Try the two ways:

s1[n_] := Sum[ (-1)^((k - 1)/2)
Sum[ 1/k^m/(m^2 + 2 m), {m, 1, Infinity, 2}], {k, 1, n, 2}];
s2[n_] := Sum[ Sum[ (-1)^((m - 1)/2)/m^k,
{m, 1, Infinity, 2}] / (k^2 + 2 k), {k, 1, n, 2}];


where s1[n] > s2[n]. Notice also the the code

s2[Infinity] // FunctionExpand // Simplify // InputForm


returns the expression (3*Zeta[3])/(2*Pi^2) which is wrong as you pointed out.

Of course, you are also free to use a modified definition of $$\beta()$$. Try this

beta1[k_?OddQ] := Sum[(-1)^((n - 1)/2)/n^k, {n, 1, Infinity, 2}];
beta2[k_?OddQ] := If[k == 1, Pi/4, (Zeta[k, 1/4] - Zeta[k, 3/4])/4^k];


They both give the same results for odd positive integers.

• Thanks: See my update at the bottom of my question. Solved as you suggested by splitting the summation into two parts. Apr 11, 2019 at 0:26
• It should perhaps be noted that DirichletBeta[] is built-in. Jun 8, 2020 at 6:14