I am trying to show that $\sum_{n=0}^{\infty}$ $\frac{x^{2n+1}}{2n+1}$ $-$ $\frac{x^{n+1}}{2n+2}$ converges pointwise but not uniformly on $[0,1]$.
My current technique is the split the domain into 3 pieces to show pointwise convergence: 1) $x=0$ 2) $x=1$ and 3) $x\in(0,1)$.
I get stuck on the case for $x\in(0,1)$.
Also I am trying to find a value in which the series "blows up" in $[0,1]$ therefore, showing the series does not converge uniformly but can't.
If anyone could provide a hint or insight that would be great.
For $0\leq x<1$, $$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}=\text{tanh}^{-1} x$$ (the inverse of the hyperbolic tangent). and $$\sum_{n=0}^{\infty}\frac{x^{n+1}}{2n+2}=-\frac{1}{2}\log(1-x)$$ hence the sum of the series, for $0\leq x<1$, is $\frac{1}{2}(2\text{tanh}^{-1}x+\log(1-x))$, and the limit of this expression as $x\to 1$ is $\frac{1}{2}\log(2)$. If the convergence was uniform, the sum would be a continuous function in $[0,1]$. In particular, the value of the series at $x=1$ would have to be $\frac{1}{2}\log(2)$. But $$\sum_{n=0}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+2}=\log(2)\neq\frac{1}{2}\log (2)$$ so the convergence is not uniform.