# Amateur Power Series question

I am almost embarrassed writing this. But can someone tell me why this may not be true (so, please give me a counter example) for a power series where $x \in [-1,1]$

$|\sum_{n \geq 0} a_n x^n| \leq \sum_{n \geq 0} a_n$

Where $\sum_{n\geq 0} a_n$ is known to be convergent.

What if $a_n \geq 0$? As an added condition, great answers so far, thanks

How about $a_n=\frac{(-1)^n}{n}$ then at $x=-1$ the sum on the left diverges and the sum on the right is $log(2)$.

• Oh great someone else had the same idea, I'll delete mine, though n is greater than or equal to 0, so...
– jgon
Dec 8, 2013 at 7:17

Take $a_0=1, a_1=-1, x=-1$ and all other $a_k = 0$. Then $|\sum_n a_nx^n| = |1.1+(-1)(-1)| = 2$, but $\sum_n a_n = 1+(-1) = 0$.

If $a_n \ge 0$, and $|x| \le 1$, then clearly $|\sum_a a_n x^n| \le \sum_n a_n |x^n| \le \sum_n a_n$.

• Thanks; as another add-on, if $a_n$ is bounded, then the result still follows. A weaker condition. Dec 8, 2013 at 7:40
• Well, the left hand side may be meaningless if $\sum_n a_n$ is not convergent. Dec 8, 2013 at 7:43

Take all the $a_n$'s to be negative, say $a_n=-1/n^2$. Then $\sum_{n}a_n$ converges. However the right side of the inequality is negative while the left side of the inequality has to be non-negative.