# Expression of the infinite summation result of a series

The infinite sum of a series should be

$$\sum_{n=1}^{\infty} \frac{x^{4 n+1}}{4 n+1}$$ $$=\frac{1}{4} \ln \frac{1+x}{1-x}+\frac{1}{2} \arctan x-x \quad(-1

However,

Sum[x^(4*n + 1)/(4*n + 1), {n, 1, Infinity}]


x (-1 + Hypergeometric2F1[1/4, 1, 5/4, x^4])

How to get the result: $$\frac{1}{4} \ln \frac{1+x}{1-x}+\frac{1}{2} \arctan x-x \quad$$ ?

Mathematica v12.2 FunctionExpand evaluates

sol=Sum[x^(4*n + 1)/(4*n + 1), {n, 1, Infinity}] // FunctionExpand//Expand
(*-x + ArcTan[x]/2 + ArcTanh[x]/2*)


If necessary ArcTanh[] might be transformed too

sol  /. ArcTanh[x] -> (ArcTanh[x] // TrigToExp)
(*-x + ArcTan[x]/2 + 1/2 (-(1/2) Log[1 - x] + 1/2 Log[1 + x])*)

• Great! Thanks a lot! Commented Mar 15, 2022 at 9:34

Another way is as follows. First, we differentiate the sum under consideration

Sum[D[t^(4*n + 1)/(4*n + 1), t], {n, 1, Infinity}]


-(t^4/(-1 + t^4))

Now we integrate the result from 0 to x

Integrate[-(t^4/(-1 + t^4)), {t, 0, x}, Assumptions -> x > -1 && x < 1]


ConditionalExpression[-x + ArcTan[x]/2 + ArcTanh[x]/2, x >= 0]

I leave math ground on your own.

• Thank you very much! Commented Mar 15, 2022 at 11:18