Taylor series of $\ln(1+x)$ So let's say we want to obtain the Taylor series for $\ln(1+x)$. We know that its derivative is $\dfrac{1}{1+x}$, which has the series $\sum_{n=0}^{\infty} (-1)^nx^n$. Can we take the antiderivative of this sum (i.e. $\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{n+1}) $ to obtain the series for $\ln(1+x)$?
 A: Yes  we can integrate term by term a power series on its domain of convergence so in your case 
$$\sum_{n=0}^\infty (-1)^n x^n$$
is a power series and its domain of convergence is $(-1,1)$ so for all $x\in(-1,1)$ we have
$$\ln(1+x)=\int_0^x\frac{dt}{t+1}=\int_0^x\sum_{n=0}^\infty (-1)^n t^ndt=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}$$
A: Yeas, that's a good approach in this case.
In fact if we know $$F'(x)=f(x)$$ in some neighbourhood of $x=x_0$ and that $f(x)$ has Taylor series
$$f(x)=\displaystyle\sum_{i=0}^{p}\frac{1}{i!} (x-x_0)^i \frac{d^if}{dx^i}(x_0)+o((x-x_0)^p)$$
Then it yields that
$$F(x)=F(x_0)+\displaystyle \sum_{i=0}^{p}\frac{1}{(i+1)!} (x-x_0)^{i+1} \frac{d^if}{dx^i}(x_0)+o((x-x_0)^{p+1})$$
A: The power series
$$
\sum_{n\ge0}a_n x^n
$$
and
$$
\sum_{n\ge0}a_n\frac{x^{n+1}}{n+1}=x\sum_{n\ge0}a_n\frac{x^n}{n+1}
$$
have the same radius of convergence, because
$$
\limsup_{n\to\infty}\sqrt[n]{\frac{|a_n|}{n+1}}=\limsup_{n\to\infty}\sqrt[n]{|a_n|}
$$
and $\lim_{n\to\infty}\sqrt[n]{n+1\mathstrut}=1$.
A bit more complicated is showing that a series and its term by term derivative have the same radius of convergence; the two results, put together, say that the term by term derivative of a power series is the derivative of the function (in the open interval of convergence). The character of a power series and its antiderivative/derivative may differ on the boundary of the interval of convergence.
So $\log(1+x)$ and $\sum_{n\ge1}(-1)^{n+1} x^n/n$ have the same derivative on $(-1,1)$. Hence, on this interval, they differ by a constant, which is easily seen to be $0$.
