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

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?

• The convergence is uniform, by the Weierstrass M-test. I don't see that Weierstrass's approximation theorem is useful here. Nov 30, 2018 at 4:53

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.

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.