# Proving Remainder of Taylor Series of 1/(1-x) approaches 0

It is well known that the Taylor (Maclaurin) series of $f(x) = \frac{1}{1-x}$ is $\sum_{n=0}^\infty x^n$ on $(-1,1)$. I am having difficulty proving the equality of these two.

The error term is bounded: $\big|R_n(x)\big| \leq \frac{\max\left|\,f^{(n+1)}(z)\right|}{(n+1)!}\big|x^{n+1}\big|$, where $|f^{(n+1)}(z)| = \frac{(n+1)!}{(1-z)^{n+2}}$.

When $-1<x<0$, I can show $\lim_{n\to\infty}R_n(x) \to 0$. I have trouble when $0<x<1$:

For a fixed $x$, $\max|f^{(n+1)}(z)| = \frac{(n+1)!}{(1-x)^{n+2}}$. Then:

$$\big|R_n(x)\big| \leq \frac{\max\left|\,f^{(n+1)}(z)\right|}{(n+1)!}\big|x^{n+1}\big| \leq \frac{(n+1)!/(1-x)^{n+2}}{(n+1)!}x^{n+1} = \frac{x^{n+1}}{(1-x)^{n+2}}.$$ As $n\to\infty$, this does not approach 0 when $x$ is near $1$.

Can someone point out either the error in my logic or a better way to bound $f^{(n+1)}(z)$?

Yes, bounding the remainder in that way only takes you to $\frac{1}{2}$, beyond that, the bound does not converge to $0$ anymore.

We can use the integral form of the remainder term,

$$R_n(x) = \frac{1}{n!} \int_0^x (x-t)^n f^{(n+1)}(t)\,dt,$$

to obtain the convergence. For $0 < x < 1$, we have (since everything is positive)

$$0 < R_n(x) = (n+1)\int_0^x \frac{(x-t)^n}{(1-t)^{n+2}}\,dt < \frac{n+1}{(1-x)^2} \int_0^x \left(\frac{x-t}{1-t}\right)^n\,dt,$$

and since $\frac{x-t}{1-t} \leqslant x$ for $0 \leqslant t \leqslant x$, the bound

$$\lvert R_n(x)\rvert \leqslant \frac{(n+1)x^{n+1}}{(1-x)^2}$$

follows, and that bound converges to $0$ for $n\to\infty$.