Proving $f(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ... $ is continuous for a fixed $x_0 \in (-1,1)$ by using the Weierstrass M-test

Question

I am trying to prove that $f(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$ is continuous for a fixed $x_0 \in (-1,1)$ by using the Weierstrass M-test. Now, the solution in the book is not what I expected, and not even close to what I had in my own mind. So, can some one verify if I am totally of with my approach to solve this

Here was my suggestion:

Let $M_n = x^n$, then for each $f_n(x) = \frac{x^n}{n}$ we have that $f_n(x) \leq M_n$. We know that $\sum_{n=1}^{\infty} M_n$ converges since it is a geometric series, hence $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly by Weierstrass M-test, and by the Continuous Limit Theorem f(x) is continuous at $x_0$.

I guess there is a mistake here somewhere...could anybody point it out for me?

Answer 1

I really think you can prove it like this: just choose $a \in (-1, 1)$ such that $x_0 \in (-a, a)$. Then for your $f_n$ the inequality $\lvert f_n(x) \rvert \leq \frac{a^n}{n} =: M_n$ holds for all $x \in [-a, a]$. Obviously $\displaystyle \sum_{n = 1}^{\infty} M_n < \infty$ is true (just estimate it by a geometric series). Hence $\displaystyle \sum_{k = 1}^n f_n(x)$ converges uniformly on $[-a, a]$, i.e. $f \lvert_{[-a, a]}$ is continuous. But as $x_0 \in [-a, a]$, this is all we wanted.

Answer 2

The condition for continuity can be written

$$\lim_{x\to x_0}\sum_{k=1}^\infty \frac{x^k}{k}=\sum_{k=1}^\infty \frac{x_0^k}{k}.$$

As the series converge absolutely, we can subtract term-wise and

$$\lim_{x\to x_0}\sum_{k=1}^\infty \frac{x^k-x_0^k}{k}=\lim_{x\to x_0}(x-x_0)\sum_{k=1}^\infty \frac1k\sum_{i=0}^{k-1}x^ix_0^{k-1-i} \\\le\lim_{x\to x_0}(x-x_0)\sum_{k=1}^\infty \frac1k\sum_{i=0}^{k-1}x_1^k =\lim_{x\to x_0}(x-x_0)\sum_{k=1}^\infty x_1^k \\=\lim_{x\to x_0}\frac{x-x_0}{1-x_1}=0$$ where we have chosen a neighborhood of $x_0$ such that $x,x_0<x_1<1$.