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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$\begingroup$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 $x=2$.
$\endgroup$ $\begingroup$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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