Does $\sum\frac1{n^2}x^n$ represent a continuous functions on $[-1,1]$?

Here is what I thought:
Let $g_n(x)=\frac1{n^2}x^n$. Since each function $g_n$ is continuous on $[-1,1]$, the infinite series $\sum g_n$ represents a continuous function if on $[-1,1]$ if this series converges uniformly on $[-1,1]$.

So I need to prove that this series converges uniformly on $[-1,1]$. I was thinking that I can show this by the following reasoning:

Let $M_n=\frac1{n^2}$. Then $\sum M_k<\infty$.
Let $x\in [-1,1] \implies |x|<1 \implies |x|^n<1\implies |x^n|<1 \implies |\frac{x^n}{n^2}|<\frac1{n^2}$.
Therefore $|g_n(x)|\leq M_k$ for all $x\in [-1,1]$. And so the series converges uniformly by the Weierstrass M-test.

This is how I would prove this myself, but my solution manual introduces a fixed number $a$, and I don't know why they do this, and if this is necessary for the proof:

This is a series which converges at both $x=-1$ (by the alternating series test) and at $x = 1$ (convergent p-series). Now consider the interval $-1\leq a\leq 1$ and note that $\sum\frac1{n^2}a^n$ converges. Since $|n^{-2}x^n|\geq|n^{-2}a^n| = > \left(\frac{a^n}{n^2}\right)$ for $x \in [-a,a]$, the Weierstrass M-test shows that the series $\sum\frac1{n^2}x^n$ converges uniformly to a function on $[-a, a]$. Since $|a|$ can be any number $\leq1$, we conclude that $f$ represents a continuous function on $[-1, 1]$.


The proof you give is fine, and the one I would use.

The only reason that I can imagine that the manual gives the other proof introducing $a$ is because it is similar the the proof of a more general principle. Namely, if a power series $\displaystyle \sum_{i=0}^{\infty}a_nx^n$ has radius of convergence $R$, then the series is uniformly convergent on $[-r,r]$ for any $r \in (0,R)$.

  • $\begingroup$ Thanks for your explanation :) $\endgroup$ – Kasper Feb 27 '13 at 13:32
  • $\begingroup$ @Kasper No worries, happy to help! $\endgroup$ – Tom Oldfield Feb 27 '13 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.