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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 $\ln(1+x)$?

Surb
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PinnaHiddda
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    Only once you have proven this is legit, i.e. for instance that it satisfies uniform convergence (it can be argued more simply using theorems about power series). In short: you can indeed, but you have to explain why you can. – Clement C. Jan 20 '15 at 16:18
  • You actually do get it this way for this problem: In general you need to be careful- for power series, things work out.. not so much for general series. – voldemort Jan 20 '15 at 16:18
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    This question is similar to: Taylor series of $\ln(1+x)$?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Martin Brandenburg Dec 25 '24 at 13:24

3 Answers3

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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

$$\ln(1+x)=\int_0^x\frac{dt}{t+1}=\int_0^x\sum_{n=0}^\infty (-1)^n t^ndt=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}$$

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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 $$F(x)=F(x_0)+\displaystyle \sum_{i=0}^{p}\frac{1}{(i+1)!} (x-x_0)^{i+1} \frac{d^if}{dx^i}(x_0)+o((x-x_0)^{p+1})$$

Vim
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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 $0$.

egreg
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