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I am working through a calc book and one of the problems asks the above question. However, taylor and maclaurin series have not been introduced yet.

In some worked examples, they leverage old series, playing it a bit fast and loose with interchanging summation and integration. I think everything is ok as long as you integrate for $x$ in the radius of convergence, where the polynomials that are the partial sums of the taylor series approximate the function uniformly, is that correct?

Following their examples I found a power series for $\ln(x)$ as follows: $$ \frac{1}{x}=\frac{1}{1+(x-1)}=\sum_{n=0}^\infty (-1)^n(x-1)^n $$ provided $|x-1|<1$. Then $$ \ln(x)=\int \frac{1}{x}\mathrm dx=\int \sum_{n=0}^\infty (-1)^n(x-1)^n\mathrm dx\\ = \sum_{n=0}^\infty \int(-1)^n(x-1)^n\mathrm dx=\sum_{n=0}^\infty (-1)^n\frac{(x-1)^{n+1}}{n+1}+c=\sum_{n=1}^\infty (-1)^{n-1}\frac{(x-1)^{n}}{n}+c $$ With boundary conditions yielding that $c=0$ since $$ \ln(1)=0=\sum_{n=1}^\infty 0+c=c $$ Modifying this series to be centered at $2$ gives me: $$ \sum_{n=1}^\infty (-1)^{n-1}\frac{(x-2)^{n}}{n} $$ But this is not correct. The book gives the answer as $$ \ln(2)+\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n2^n}(x-2)^n $$ Any help would be appreciated. I am especially confused about where the $\ln(2)$ term comes from if you don't know taylor series yet.

edit: follow up question, is it correct to say that the geometric series is a a maclaurin series for $f(x)=\frac{1}{1-x}$?

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    $\begingroup$ The modification you did gave you the expansion of $\ln(x - 1)$ $\endgroup$ – user258700 Jun 29 '16 at 1:42
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    $\begingroup$ @qbert: Yes, the usual $1+x+x^2+x^3+\cdots$ is the Maclaurin series for $\frac{1}{1-x}$. $\endgroup$ – André Nicolas Jun 29 '16 at 1:54
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    $\begingroup$ @qbert: The "uniform convergence" remark is not quite right, but it is true that if $R$ is the radius of convergence, then we can differentiate, integrate freely when $|x-a|\lt R$. That is a very nice feature of power series. With other types of series, including the very important trigonometric series, one has to be more careful. $\endgroup$ – André Nicolas Jun 29 '16 at 2:09
  • $\begingroup$ @AndréNicolas hm ok, that makes sense. Is there a name for the proof of that result? Specifically the antidifferentiation part. Googling around just seems to give me the method. $\endgroup$ – qbert Jun 29 '16 at 2:13
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    $\begingroup$ @qbert: I do not know of a name for the result. It is in "every" introduction to analysis book, for example Rudin. By the way, in general term by term integration is better behaved than term by term differentiation. $\endgroup$ – André Nicolas Jun 29 '16 at 2:17
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You just need to be careful about the definite integral. So, $$ \frac{1}{x}=\frac{1}{2+(x-2)}=\frac{1}{2}\frac{1}{1+\frac{1}{2}(x-2)}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(x-2)^n}{2^n}, $$ and $$ \log x - \log a=\int_{y=a}^{x}\frac{dy}{y}=\int_{y=a}^{x}\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(y-2)^n}{2^n}dy=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(y-2)^{n+1}}{(n+1)2^n}\bigg\vert_{y=a}^{x}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(x-2)^{n}-(-1)^{n-1}(a-2)^{n}}{n2^n}. $$ To make this come out nicely, take $a=2$. Then the lower limit of each term vanishes, leaving $$ \log x = \log 2 + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(x-2)^{n}}{n2^n}. $$

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  • $\begingroup$ ah! now I understand why Andre picked $(x-2)$. This is clever. $\endgroup$ – qbert Jun 29 '16 at 1:48
  • $\begingroup$ Also, maybe this doesn't matter, but why wasn't the $2^n$ re-indexed? $\endgroup$ – qbert Jun 29 '16 at 2:21
  • $\begingroup$ It was... it went to $2^{n-1}$ when the sum was reindexed, and then I pulled in the factor of $1/2$ to make it $2^{n}$ again. $\endgroup$ – mjqxxxx Jun 29 '16 at 2:34
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Hint: Note that $$\frac{1}{x}=\frac{1}{2+(x-2)}=\frac{1/2}{1+(x-2)/2}.$$ Now expand as you know how to do, and integrate term by term. Choose the constant term suitably to give you the right answer at $x=2$.

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