What is "wrong" with $\sum_{n=0}^\infty \frac{(-2x)^n}{1+n}$ at $x = 0$ 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.
 A: 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.
A: The sum can be evaluated at $0$ by L'Hospital Rule:
$$
\lim_{x\to0}\frac{\ln(1+2x)}{2x}=\lim_{x\to0}\frac{1}{1+2x}=1
$$
A: For $|x|<\frac{1}{2}$, the sum
$$\sum_{n=0}^\infty \frac{(-2x)^n}{1+n}$$
converges to
$$
g(x)=
\begin{cases}
 \frac{\log(1+2x)}{2x},&0<|x|<\frac12; \\
1,&x=0.
\end{cases}
$$
A: 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.
