# For $\sum_{0}^{\infty}a_nx^n$ with an infinite radius of convergence, $a_n \neq0$ , the series does not converge uniformly in $(-\infty , \infty)$.

I'd like your help with the following question:

Let $\sum_{0}^{\infty}a_nx^n$ be a power series with an infinite radius of convergence, such that $a_n \neq 0$ , for infinitely many values of $n$. I need to show that the series does not converge uniformly in $(-\infty , \infty)$.

I quite don't understand the question, since I know that within the radius of convergence for a power series, there's uniform convergence, and since I know that the radius is "infinite", it says that the uniform convergence is infinite, no?

Thanks!

• The theorem you are referring to is that you have uniform convergence on $[-a,a]$ for any real number $a$ with $a<R$, where $R$ is the radius of convergence. Commented Feb 11, 2012 at 16:21

Hint: put $s_N(x):=\sum_{n=0}^Na_nx^n$. If the sequence $\{s_N\}$ is uniformly convergent on $\mathbb R$ then the sequence $\{s_{N+1}-s_N\}$ converges uniformly on $\mathbb R$ to $0$ so $\{a_{N+1}x^{N+1}\}$ converges uniformly to $0$. Do you see the contradiction?

• Do you use Cauchy and want to show that $|\sum_{n=0}^Na_nx^n-\sum_{n=0}^{N+1}a_{n+1}x^{n+1}| \geq \epsilon$? Commented Feb 11, 2012 at 16:26
• It's like Cauchy since we have denoting $f$ the power series $||s_{N+1}-s_N||\leq ||s_{N+1}-f||+||f-s_n||$ which converges to $0$. In particular for $n$ large enough $\sup_{x\in\mathbb R}|a_nx^n|\leq 1$. Commented Feb 11, 2012 at 16:30
• If you don't mind going step by step, I will really appreciate it, cause I got confused. "If the sequence $\{s_N\}$ is uniformly convergent on $\mathbb R$ then the sequence $\{s_{N+1}-s_N\}$ converges uniformly on $\mathbb R$"- This is Cauchy, right? Commented Feb 11, 2012 at 16:36
• Yes, it's a Cauchy sequence and you can see that by triangular inequality. Commented Feb 11, 2012 at 16:37
• As explained by @Davide, this is not Cauchy but rather the general fact that if $u_n\to u$ then $u_{n+1}-u_n\to0$.
– Did
Commented Feb 11, 2012 at 18:21

First, your argument is flawed. I assume the theorem you are referring to is that you have uniform convergence on $[−a,a]$ for any real number $a$ with $a<R$, where $R$ is the radius of convergence of the power series.

But you can not take $a=\infty$ in your problem...

What you could use is the following, very useful, fact:

A series $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$ if and only if the sequence of partial sums $S_n=\sum\limits_{m=1}^n f_m$ is uniformly Cauchy. That is, for every $\epsilon>0$, there is an integer $N$ such that $$|S_m(x)-S_n(x)|=\Bigl|\sum_{k=n+1}^m f_k(x)\Bigr|<\epsilon,\quad\text{ for all }x\in I\ \ \text{and all }m\ge n \ge N.$$

An immediate result of this (and a result that, by itself, is easily proven, as in Davide Giraudo's answer) is:

If $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$, then for every $\epsilon>0$, there is an integer $N$ such that $$|f_n(x)| =| S_{n }(x)-S_{n-1}(x)|<\epsilon,\quad\text{ for all }x\in I\ \ \text{ and all }n \ge N;$$ that is, the sequence $(f_n)$ converges uniformly to the zero function on $I$.

One can use the first result I mentioned directly to show that your series is not uniformly Cauchy over $\Bbb R$:

Since $a_i\ne 0$ for infinitely many $n$, given any positive integer $N$, you can choose $m>n\ge N$ such that$\sum\limits_{k=n}^m a_k x^k$ is not a constant polynomial. Taking the limit as $x$ tends to infinity will verify that this sum is not uniformly close to 0.

Let $X$ be a set. If $(f_n)$ is a sequence of functions from $X$ to $\mathbb C$, and if the series $\sum_{n\ge0}f_n$ converges uniformly on $X$, then the sequence $(f_n)$ converges uniformly to $0$.

Proof: Write $$f_n=\sum_{k=0}^n\ f_k-\sum_{k=0}^{n-1}\ f_k$$ and let $n$ tend to $\infty$.