Dr. E: Mathematica says In[43]:= Sum[(-1)^n x^(3*n)/(3*n)!, {n, 0, \[Infinity]}] Out[43]= 1/3 E^-x (1 + 2 E^(3 x/2) Cos[(Sqrt[3] x)/2]) n is used twice. In[45]:= Reduce[%43 == 0] Out[45]= E^(x/2) != 0 && Cos[(Sqrt[3] x)/2] != 0 && E^(3 x/2) == -(1/2) Sec[(Sqrt[3] x)/2] Looks too hard for a Lambert-W. Beats me. --Bill On Sat, Mar 10, 2018 at 2:50 PM, françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
Hello
In an unfortunate attempt to solve zeta (2n + 1), I became interested in certain functions.
I was able to find:
cos (x / sqrt (2)) * cosh (x / sqrt (2)) = sum ((- 1) ^ n * x ^ (4 * n) / (4 * n)!,n=0..infinity)
but I am lost to the next sum.
sum ((- 1) ^ n * x ^ (3 * n) / (3 * n)!,n=0..infinity) = ???
Do you know the function returned by this sum?
I just noticed that the solutions of the equation
sum ((- 1) ^ n * x ^ (3 * n) / (3 * n)!,n=0..infinity) = 0
are substantially close to:
(2 * n + 1) * Pi / sqrt (3)
Is there more information on this?