let all xn>=0 for all natural numbers n. Assume that sigma(n from 1 to infinity) xn<infinity
Prove that the lim as n tends to infinity of xn=0
Assume not; that is, $\displaystyle x_n\to L>0$. Because $\displaystyle x_n$ converges, $\displaystyle \exists N$ such that $\displaystyle n\geq N$ implies $\displaystyle |x_n-L|<\frac{L}{2}\implies \frac{L}{2}<x_n<\frac{3L}{2}$
Let $\displaystyle \sum_{n=1}^{N-1} x_n=K$ (because after all a finite sum is just a number).
Now we have $\displaystyle \sum_{n=1}^{\infty}x_n=K+\sum_{n=N}^{\infty}x_n>K+ \sum_{n=N}^{\infty}\frac{L}{2}$
So...
Yes. Since $\displaystyle L$ is a constant, you can pull it out of the sum.
$\displaystyle \sum_{n=N}^{\infty}\frac{L}{2}=\frac{L}{2}\sum_{n= N}^{\infty}1$
which clearly diverges to $\displaystyle \infty$. This is a contradiction, so $\displaystyle L=0$.
I just wanted to show another way to do this.
Let $\displaystyle \sum_{n=1}^{\infty}x_n=K$
You know that $\displaystyle \sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n = x_N$
So taking the limit:
$\displaystyle \lim_{N\to\infty}x_N=\lim_{N\to\infty}\sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n$
But because you know that the sum is finite:
$\displaystyle =K-K=0$
Yet another way: Let $\displaystyle S_n= \sum_{n=1}^N a_n$ be the nth partial sum. Then the series converges if and only if the sequence of partial sums converges (that's the definition of convergence of a sequence).
But then, if $\displaystyle a_n$ does not go to 0, $\displaystyle S_{n+1}- S_n= a_{n+1}$ does not go to 0. That means that $\displaystyle S_{n}- S_m$ cannot go to 0 as n and m go to 0 independently and so $\displaystyle {S_n}$ is not a Cauchy sequence.
Since every convergent sequence is a Cauchy sequence, $\displaystyle {S_n}$ does not converge and thus $\displaystyle \sum_{n=1}^\infty a_n$ does not converge.