Divergence of a Taylor series

We showed in class that the Taylor series of $e^x$ converges everywhere to $e^x$ and you have (or will) establish a similar result for $\sin(x)$. Where, exactly, does this argument fail for
$$f(x)=\frac{1}{1-x}?$$

Note that $$\frac{1}{1 - x} = \sum_{k = 0}^{\infty} x^{k}$$ $\forall$ $x\in(-1,1)$, but for $x\notin(-1,1)$, $$\sum_{k = 0}^{\infty}\left| x^{k}\right| = \infty$$

@Professor.Membrane Agreed! However, the question asks where the remainder theorem fails.

Reading more carefully would be helpful...

$$R_{n}(x) = \left|\frac{1}{1 - x} - \sum_{k = 0}^{n}x^{k}\right| = \left|\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}\right| \textrm{ for } \xi \in (0, x)$$

but unlike $e^{x}$ whose derrivative is itself; and sin$(x)$ whose derivatives are cyclic and bounded by $1$, The derivatives of $f$ bounce around all over the place. So it seems like there is no way to show that $R_{n}(x) \rightarrow 0$ as $n \rightarrow \infty$.

@Professor.Membrane I think you can, in fact, find a reasonably simple formula for $f^n(0)$. Using that, you should be able to see that
$$R_n(x) \to 0$$
for $x\in(-1,1)$, but not for $x\geq1$.

@mark

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

look better?

$$f^{n}(0) = (-1)^{n}(n - 1)!$$

Then $$ R_{n}(x) = \left|\frac{(-1)^{n}(n - 1)!}{(n + 1)!}x^{n+1}\right| = \left|\frac{x^{n + 1}}{n ^{2} + n}\right|$$

Which pretty clearly converges to $0$ is $x\in(-1,1)$ and otherwise diverges.

Maybe I am just not seeing it but how does this implication shows it does not work. Again this might just be me.

@Ricky_Bobby and @Professor.Membrane

I think that Ricky has a valid point. I'm not convinced that your formula for $f^n(0)$ is correct. If it were, then it seems to me that the remainder would approach zero for all $x$.

Keep in mind - you're going to start with $f(x) = (1-x)^{-1}$. Then, you're going to repeatedly differentiate. I suspect that the formula for $f^{(n)}(x)$ is going to involve $n!$.

Also, if you do add more, then it's best to edit your existing response, as opposed to adding another response.