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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 $[-1, 1]$.

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1 Answer 1

up vote 2 down vote accepted

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 $r \in (0,R)$.

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

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