Solving an infinite sum of tanh(i*pi/2)/i^5

Questions
1

Hi! I’m stuck with evaluating the following sum:

Sum[Tanh[(Pi*i/2)/i^5], {i, 1, Infinity}]

I thought of solving it using the Riemann Zeta functions but didn’t quite get it and and wolfram provides only a numerical value. How would you go about solving it step by step? Thank you!

2

I couln’t go beyond this

Map[Sum[#, {i, 1, Infinity}] &,
 Tanh[Pi*i/2]/i^5 // TrigToExp // Apart]
3

I think that does not have finite closed-form expression in terms of very large class of special functions.

\sum _{i=1}^{\infty } \frac{\tanh \left(\frac{i \pi }{2}\right)}{i^5}=\\\zeta (5)-2 \sum _{k=1}^{\infty } \text{Li}_5\left(e^{-k \pi }\right)+4 \sum _{k=1}^{\infty } \text{Li}_5\left(e^{-2 k \pi }\right)=\\\zeta (5)-2* \text{A255701}+4* \text{A255702}

Where A255701 and A255702

  {NSum[Tanh[(i \[Pi])/2]/i^5, {i, 1, Infinity}, 
    WorkingPrecision -> 31], 
   Zeta[5] - 
    N[Sum[2 PolyLog[5, E^(-k \[Pi])] - 4 PolyLog[5, E^(-2 k \[Pi])], {k,
        1, 1000}], 21]}

 (*{0.9539629221432281032692, 0.9539629221432281032692}*)

Regards M.I.