Does the convergence of $\sum_{n = 0}^{\infty} |a_{n}| \Rightarrow |a_{n}| \rightarrow 0$ Consider the power series $\sum_{n = 0}^{\infty} a_{n} x^{n}$.
If $\sum_{n = 0}^{\infty} |a_{n}|, n \in \mathbb{Z}$ is absolutely convergent, does that necessarily imply $|a_{n}| \rightarrow 0$? 
When I think about it, it seems true. I'm not asking for a proof or anything I just want to get an intuition of why that is true (if it is to begin with).
 A: $$s_{n}=|a_{1}|+|a_{2}|+...+|a_{n-1}|+|a_{n}|\\\sum_{n=0}^{\infty }|a_{n}|=M\\\lim_{n \to \infty }s_{n}=M\\\lim_{n \to \infty }|a_{1}|+|a_{2}|+...+|a_{n-1}|+|a_{n}|=M\\\lim_{n \to \infty }s_{n-1}+|a_{n}|=M\\$$when $$n \to \infty \\\lim_{n \to \infty }s_{n-1}=\lim_{n \to \infty }s_{n}=M\\$$so $$\lim_{n \to \infty }s_{n-1}+|a_{n}|=M\\\lim_{n \to \infty }s_{n-1}+\lim_{n \to \infty }|a_{n}|=M\\M+ \lim_{n \to \infty }|a_{n}|=M\\\lim_{n \to \infty }|a_{n}| \to 0$$ 
A: Suppose $|a_n|$ converged to some nonzero number $L>0$. That implies that for any $\epsilon>0$, there is a natural number $N$ such that $L-\epsilon<|a_n|$ for all $n\ge N$. But then we would have
$$\sum_{n=0}^{\infty}|a_n|\ge \sum_{n=N}^{\infty}|a_n|>\sum_{n=N}^{\infty}(L-\epsilon)=\infty.$$
Contradiction.
A: An intuitive explanation of why it needs to be true can be found in looking at the contrapositive, i.e.
$$
\lvert a_n \rvert \not\to 0 \implies \sum_{n = 0}^\infty \lvert a_n \rvert \text{ is not convergent}
$$
If we have the case where $\lvert a_n \rvert \not\to 0$ then it either converges to some positive number, call it $a$, diverges or doesn't converge (in the case of diverging or simply not converging I encourage you to fill in the details). If it converges, for large enough $n$ the terms $\lvert a_n \rvert$ are larger than $a / 2$, so that we have
$$
\sum_{n = N}^\infty \lvert a_n \rvert > \sum_{n = N}^\infty \frac{a}{2} = \infty
$$
The point here is if it converges to finite some value, there must be some point where all the terms are large enough that if you add them up they get unboundedly large.
A: You "just want to get an intuition of why that is true" OK so it's simply :
Imagine you have to sum an infinite numbers of positive numbers which have 100 for sum you can do like this : 
First you have the sum to 90 : $1+1+1+1+2+4+10+20+30+15+5=90=A$
Here you just have finite sum numbers so you add more terms $99=A+8+1$
If you want obtain one you can doing $1=1/2+1/2$ but it's not an infinite sum so you have to add some numbers but more and more small :) 
$1=1/3+1/4+1/8+1/14+1/48+1/49+1/50+1/170+1/185+ ...+\epsilon$ with   necessarily $\epsilon \rightarrow 0 $ 
Shadock 
