# Prove that the series $\sum_{n=0}^{\infty}$ $\frac{x^{2n+1}}{2n+1}$ $-$ $\frac{x^{n+1}}{2n+2}$ converges pointwise but not uniformly on $[0,1]$

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.

• Two points: For $x$ in the interior, ratio test works fine for pointwise convergence. The non-uniform situation is around $x=1$. – herb steinberg Apr 13 at 21:20
• If it helps at $x=1$ you get the alternating harmonic series, which is none other than $\log(2).$ – Melody Apr 13 at 21:31
• @Matthieu well do you know the result that if continuous functions converge uniformly, then their limit is continuous? This can be helpful to get an intuition. Like, let's look at the sequence $x^n$ defined on $[0,1].$ You can show this converges uniformly to $g$ where $g(x)=0$ on $[0,1)$ and $g(1)=1.$ Since $g$ is not continuous the convergence there is not uniform. So if we can show any sort of discontinuity in a pointwise limit, then we don't have uniform convergence. The same goes for series. – Melody Apr 13 at 21:50
• @Melody Great! that helps a lot thank you. Also did you mean to write "You can show this converges uniformly to $g$..."? The reason I ask is because if this were the case, as you said, the limiting function should be continuous. – Matthieu Apr 13 at 21:56
• @Matthieu I didn't mean to. That was a mistake. I meant pointwise. I'd edit the comment, but editing it is disabled at the moment. So just know I meant pointwise. It is not uniform, because the lack of continuity. – Melody Apr 13 at 22:03

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.
• Could you clarify how you got $x \to 1$ is $\frac{1}{2}\log(2)$? – Matthieu Apr 13 at 22:21
• @Matthieu If you call the sum $f(x)$ then term-by-term differentiation (you allowed to do this by the uniform convergence of this series) gives that $f'(x) = \sum x^{2n} - x^n = \frac{1}{2(x+1)}$ so $\lim_{x\to 1}f(x) = \int_0^1 \frac{{\rm d}x}{2(x+1)} = \frac{\log(2)}{2}$. – Winther Apr 13 at 22:28
• @Winther -- Thank you, that's nice, but the limit is still $\frac{1}{2}\log(2)$ even without uniform convergence, which aposteriori does not hold. – uniquesolution Apr 13 at 22:29