Prove that $f(x) = \sum_{n=1}^{\infty} x^n/n^2$ is continuous on $[0,1]$

Question

I have a general idea of how to prove this but I could use some help with the details.

Basically I see that $f(x)$ is the uniform limit of $f_k(x) = \sum_{n=1}^{k} x^n/n^2$ on $[0,1]$.

Each $f_k$ is continuous, so $f$ is as well since uniform convergence preserves continuity.

Is this proof correct/does it seem sufficient?

Thanks ahead of time.

Answer 1

Note that $|x^n/n^2|\leq 1/n^2$ on $[0,1]$ and $\sum_{n=1}^{\infty} 1/n^2 \lt \infty$. Then by Weierstrass M-test,$\sum_{n=1}^{\infty} x^n/n^2$ converges uniformly. Hence $f$ is continuous.

Answer 2

This can be done by a calculus student, actually, and I can even prove the continuity is uniform, provided said calculus student recalls the mean value theorem. No $M$-test needed here!

Let $x_0, x_1 \in [0,1]$, and let $\epsilon > 0$ be given. We'll find $\delta$ so that $|x_0 - x_1| < \delta \implies |f(x_0) - f(x_1)| < \epsilon$.

To see this, introduce the auxiliary functions $g_n(x) = x^n$

We have:

$$|f(x_1) - f(x_0)| = \bigg|\sum_{n=1}^\infty \frac{x_1^n - x_0^n}{n^2} \bigg| = \bigg| \sum_{n=1}^\infty \frac{g_n(x_1) - g_n(x_0)}{n^2}\bigg|$$

Now $g_n$ are all differentiable, so $g_n(x_1) - g_n(x_0) = g_n'(x)(x_1 - x_0)$ by the Mean Value Theorem, for some $x \in (x_0, x_1)$ (assuming without loss of generality $x_1 > x_0$.

Now we have $g_n'(x) = nx^{n-1}$, so shoving this in to the above equation and observing that now all the terms are positive in the sum, so we drop the bars:

$$|f(x_1) - f(x_0)| \leq \bigg[\sum_{n=1}^\infty \frac{x^{n-1}}{n}\bigg] |x_1 -x_0|$$

But now, the first term is a convergent series, since $x < 1$, and $n > 0$, this expression is bounded above by a geometric series. Calling the limit value $S$, taking $\delta = 1/S$ suffices.