Sum of power series $\frac{x^{2n-1}}{2n-1}$

Question

Consider the power series defined by $\frac{x^{2n-1}}{2n-1}$: $\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+...$. Its radius of convergence is clearly 1, for any $x \geq 1$ diverges by comparison with the harmonic series, and any $0 \leq x < 1$ converges by comparison with a geometric series.

Is there a way to calculate its sum? What method can be used? I request that answers describe the appropriate method but do not complete the solution.

Answer 1

Hint:

If $f(x) = x+\frac {x^2}2 +\frac {x^3}3 +\frac {x^4}4 +\frac {x^5}5 +\frac {x^6}6 +\cdots$ then consider

• $\frac12 f(x^2)$
• $\frac{d}{dx} f(x)$

and see whether this suggests anything familiar

Answer 2

Let $f(x)=\sum\limits_{n=1}^\infty \frac{x^{2n-1}}{2n-1}$ Then $f'(x)=\sum\limits_{n=1}^\infty x^{2n-2}=\sum\limits_{m=0}^\infty (x^2)^m$.
Sum the geometric series and integrate to get $f(x)$.

Answer 3

We know that $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{2n-1}=\arctan x.$$ Now use the identity $\arctan (ix)\equiv i\operatorname{artanh}x$.