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Let $r>0$. I want to use the Lagrange Form of the Remainder (as stated, say, here) to prove that the Taylor series of $\log(x)$ centered at $r$ converges to $\log$ on $(\frac{r}{2}, \frac{3r}{2})$.

I've calculated the Taylor series to get $\sum_{n=0}^{\infty}\frac{(-1)^{n}(n-1)!}{(n!)r^n}(x-r)^n$ but am lost on how to do this problem. I'd appreciate some help.

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    $\begingroup$ "Lagrange" form, not language form xDDD Lagrange was a big mathematician. $\endgroup$
    – Masacroso
    Feb 3, 2017 at 7:11
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    $\begingroup$ Your $n$-th derivative term for $\log$ should have $(n-1)!$ in the numerator. Otherwise close. $\endgroup$
    – RRL
    Feb 3, 2017 at 7:14
  • $\begingroup$ @RRL Another issue with the OP's series is that inasmuch as the expansion is around $x=r$, the first term is $\log(r)$. $(-1)!$ does not even exist. $\endgroup$
    – Mark Viola
    Feb 3, 2017 at 22:35

1 Answer 1

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The Lagrange remainder is

$$R_n(x) =\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-r)^{n+1} = \frac{(-1)^{n}n!}{(n+1)!\xi^{n+1}}(x-r)^{n+1},$$

where $x, \xi \in (r/2, 3r/2).$

Hence,

$$\left|R_n(x) \right|= \frac{1}{n+1}\frac{|x-r|^{n+1}}{|\xi|^{n+1}} \leqslant \frac{1}{n+1}\frac{|r/2|^{n+1}}{|r/2|^{n+1}} = \frac{1}{n+1}.$$

Thus, the Taylor series converges since $R_n(x) \to 0$ as $n \to \infty$.

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