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I have no clue how to attempt this problem.

consider the power series: $$\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}$$ Find the closed form formula for the function $f(x)$ to which the power series converges.

user170231
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Chang
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2 Answers2

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$$f(x)=\sum_{n=1}^{\infty} (-1)^n\frac{x^{n+1}}{n+1}$$ $$f'(x)=\sum_{n=1}^{\infty} (-x)^n=\frac1{x+1}$$ $$\therefore f(x)=\int \frac1{x+1} dx=\ln (1+x)+C$$ $$f(0)=0 \rightarrow C=0$$

Kay K.
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  • Ok, why does -x^n become 1/x+1? – Chang Dec 10 '15 at 15:50
  • Generally, try to make the series to something like $\sum x^n$ or $\sum (ax)^n$ ($a$ is a real number including $-1$ in this case). If it's in the form of $\sum nx^n$ then if you integrate it you can remove $n$. If it's in the form of $\sum \dfrac{x^n}n$ then if you differentiate then you can remove $n$ too. Sometimes it's more complex form like $\sum n^2x^n$ or $\sum \dfrac{x^n}{n^2}$ then you'll have to integrate or differentiate several times. – Kay K. Dec 10 '15 at 15:57
  • You probably need to multiply or divide by $x$ or $x^m$ $(m\in\mathbb N)$ to adjust $x$'s exponent. Take a look at this and you'll see how you should apply this strategy depending on situations. http://math.stackexchange.com/questions/1566506/sum-of-binomial-coefficients-multiplied-by-k2/1566522#1566522 – Kay K. Dec 10 '15 at 15:58
  • The above applies only to the form of $\sum (ax)^n=\frac1{1-ax}$. You can also remember Taylor series expansions of some famous functions like $\sum \frac{x^n}{n!}=e^x$ and then you'll be able to solve more problems like this kind. – Kay K. Dec 10 '15 at 16:01
  • Well, to answer to why $\sum (-x)^n = \frac1{x+1}$, let $S=\sum_{n=0}^{\infty} (-x)^n$. And then try to add these two series $S+xS$ and see what you get. – Kay K. Dec 10 '15 at 16:05
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One way forward is to write for $-1<x\le 1$

$$\begin{align} \sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{n+1}&=\sum_{n=0}^\infty (-1)^n\int_0^x u^n\,du\\\\ &=\lim_{N\to \infty}\int_0^x \sum_{n=0}^N(-u)^n\,du\\\\ &=\lim_{N\to \infty}\int_0^x \left(\frac{1-(-u)^{N+1}}{1+u}\right)\,du\\\\ &=\log(1+x)-\lim_{N\to \infty}\int_0^x \left(\frac{(-u)^{N+1}}{1+u}\right)\,du \tag 1\\\\ &=\log(1+x)-\int_0^x\lim_{N\to \infty} \left(\frac{(-u)^{N+1}}{1+u}\right)\,du \tag 2\\\\ &=\log(1+x) \end{align}$$

where the Dominated Convergence Theorem justifies going from $(1)$ to $(2)$

Mark Viola
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  • Please let me know how I can improve my answer. I really want to give you the best answer I can. Happy Holidays! - Mark – Mark Viola Dec 29 '15 at 16:04