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?


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.

  • $\begingroup$ If I understand your question correctly, you already know the correct answer. And Mathematica knows $\beta(1)=\pi/4$. $\endgroup$
    – Somos
    Apr 10, 2019 at 17:45
  • $\begingroup$ @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. $\endgroup$ Apr 10, 2019 at 18:04
  • $\begingroup$ @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. $\endgroup$ Apr 10, 2019 at 18:23

1 Answer 1


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.

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

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