What is "wrong" with $\sum_{n=0}^\infty \frac{(-2x)^n}{1+n}$ at $x = 0$

Question

I was working with this power series:

$$\sum_{n=0}^\infty \frac{(-2x)^n}{1+n}$$

Which converges for $|x| < \frac{1}{2}$ according to Cauchy-Hadamard theorem. No problem so far.

Then I found its sum, that is:

$$ \frac{\log(1+2x)}{2x} $$

And here I get confused, since previously I found that the series converges for x=0, but I can not evaluate its sum at x = 0. What am I missing here?

Any help or comment is appreciated, and thanks for your time.

Answer 1

I think that outlining a similar situation will clear the fog. For any $x\in\mathbb{R}$ we have $$ \sin(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!} \tag{1}$$ hence by dividing both sides by $x$ (this is the crucial part) we may say that $$ \frac{\sin x}{x}=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!}=1-\frac{x^2}{6}+\ldots \tag{2}$$ but, wait, the LHS makes no sense at $x=0$!. The previous step is not really incorrect, it is just a small terminology abuse. The right thing to write was $$ \forall x\in\mathbb{R},\qquad \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!}=\lim_{z\to x}\frac{\sin z}{z},\tag{3}$$ "dodging" the problem of dividing by zero. But that is so boring that we often write $$ \forall x\in\mathbb{R},\qquad \frac{\sin x}{x}=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!}=1-\frac{x^2}{6}+\ldots \tag{3}$$ with the actual meaning that the LHS has to be interpreted as $1=\lim_{x\to 0}\frac{\sin x}{x}$ when $x=0$.
And you are facing just the same issue.

Answer 2

The sum can be evaluated at $0$ by L'Hospital Rule: $$ \lim_{x\to0}\frac{\ln(1+2x)}{2x}=\lim_{x\to0}\frac{1}{1+2x}=1 $$

Answer 3

For $|x|<\frac{1}{2}$, the sum

$$\sum_{n=0}^\infty \frac{(-2x)^n}{1+n}$$

converges to

$$ g(x)= \begin{cases} \frac{\log(1+2x)}{2x},&0<|x|<\frac12; \\ 1,&x=0. \end{cases} $$

Answer 4

Let's see how you probably got the sum of the series; if we multiply by $2x$, we get $$ \sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2x)^{n+1}}{n+1} $$ and, by a well known Taylor series expansion, $$ \sum_{n=0}^{\infty}\frac{(-1)^{n}(2x)^{n+1}}{n+1}=\log(1+2x) $$ So, if $f(x)$ is the sum of your series, we know that $$ 2xf(x)=\log(1+2x) $$ Now we can divide by $2x$, provided $x\ne0$, so $$ f(x)= \begin{cases} \dfrac{\log(1+2x)}{2x} & \text{if $x\ne0$} \\[6px] 1 & \text{if $x=0$} \end{cases} $$ (the sum for $x=0$ is easily computed).

No problem at all. By the way, since $f$ is differentiable on $(-1/2,1/2)$, hence continuous, we have $$ \lim_{x\to0}\frac{\log(1+2x)}{2x}=1 $$ which of course can be checked directly.