My question concerns the function $f = \frac{1}{1-x}$
The derivatives 1,2 ..(n+1) I believe are:
- $f^{(1)} = \frac{1}{(1-x)^2}$
- $f^{(2)} = \frac{2}{(1-x)^3}$
- $f^{(n+1)} = \frac{(n+1)!}{(1-x)^{n+2}}$
The taylor polynomial around $x=0$ would then be
$f(x) = \sum_{k=0}^n x^k+ R_n(f,0)$
with the lagrange closing term $R_n(f,0)= \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ and $c$ between $0$ and $x$
I hope so much is correct. Now a simple look at $\lim_{n\to \infty} x^n$ confirms the idea that the terms will go to $0$ as $n \to \infty$ if and only if $|x|\lt 1$ but somehow when I try to proof that using $\lim_{n\to \infty} |R_n|$ I get the opposite!
$|R_n(f,0)|= \lvert \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \rvert = \lvert \frac{\frac{(n+1)!}{(1-c)^{n+2}}}{(n+1)!}x^{n+1}\rvert = \lvert \frac{x^{n+1}}{(1-c)^{n+2}}\rvert$
Taking the limit here: $\lim_{n\to \infty} \lvert \frac{x^{n+1}}{(1-c)^{n+2}}\rvert = L$
Now this limit diverges for values of $|x|<1$
... I've been staring at it but did not find the mistake. Any pointers appreciated!
There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+\cdots$ converges to $\frac1{1-x}$ for $|x|<\frac12$. This is true. The fact that the series also converges to that value for $\frac12\le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.
The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.