TITLE: Simplification of Taylor Series of ln(x) about 1
QUESTION [0 upvotes]: While calculating the Taylor series of $ln(x)$ about the point $x_0=1$ I found the following:
$f(x) = \frac{1}{1!}(x-1)^1-\frac{1}{2!}(x-1)^2+\frac{2}{3!}(x-1)^3+…$
From this, I generalized the summation as $\sum\limits_{k=1}^{\infty}  (-1)^{n+1}\frac{(x-1)^n}{n!}(x-1)^n$
However, when I tried to verify my calculation on Wolfram Alpha this sum was written without the factorial in the denominator. Where exactly does the factorial go to, then?

REPLY [1 votes]: Your numerators are actually $(n-1)!$, so $(n-1)!/n!=1/n$.
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