Riemann integral of a series

Question

Let $0\leq b<1$ $$\sum_{k=1}^{\infty}\frac{x^{2k-1}}{k-\frac{1}{2}}, \quad x\in\mathbb{R}$$

• Show the the antiderivative $g:[-b,b]\rightarrow \mathbb{R}$ is differentiable and that $$g'(x)=\frac{2}{1-x^2}$$

for a $x \in [-b,b]$.

So far

I have shown that the sum converges uniformly on the interval $ [-b,b]$. After that I am completly stuck. It is my first time encountering anything quite like this. I know that if $g_n$ is a sequence of functions $g:[a,b]\rightarrow\mathbb{C}$ that are all Riemann integrable. If the sequence of functions converges uniformly to $g$, then $g$ is also Riemann integrable and

$$\lim_{n\to\infty}\int_{a}^{b}g_n(t)dt=\int_{a}^{b}g(t)dt$$

Answer 1

Consider the Taylor Series of $\ln|1+x|$: $$\ln|1+x|=\sum_{k=1}^\infty \frac {(-1)^{k+1}x^k}{k}$$ In this series, put $x\to -x$ and subtract. So, $$\ln |1-x|=-x-\frac {x^2}{2}-\frac {x^3}{3}-\frac {x^4}{4}-...$$ If you subtract, $$\ln|\frac {1+x}{1-x}|=2(x+\frac {x^3}{3}+\frac {x^5}{5}...)$$

Answer 2

If you let $$g(x) = 2\sum_{k=1}^{\infty} \frac{x^{2k-1}}{2k-1},$$ $g$ is given by a power series that converges uniformly on $[-b, b]$, thus it is infinitely differentiable.

$$g'(x) = 2\sum_{k=1}^{\infty} x^{2k-2} = 2\sum_{k=0}^{\infty}x^{2k} = \frac{2}{1-x^2}$$ using the geometric series $\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$ for $|x| < 1$.