# Limit Proof

• Nov 12th 2009, 08:25 PM
amm345
Limit Proof
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 x
n=0
• Nov 12th 2009, 09:07 PM
redsoxfan325
Quote:

Originally Posted by amm345
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 x
n=0

Assume not; that is, $x_n\to L>0$. Because $x_n$ converges, $\exists N$ such that $n\geq N$ implies $|x_n-L|<\frac{L}{2}\implies \frac{L}{2}

Let $\sum_{n=1}^{N-1} x_n=K$ (because after all a finite sum is just a number).

Now we have $\sum_{n=1}^{\infty}x_n=K+\sum_{n=N}^{\infty}x_n>K+ \sum_{n=N}^{\infty}\frac{L}{2}$

So...
• Nov 12th 2009, 09:34 PM
amm345
I'm sorry, I'm still having troubles seeing where to go next?
Am I looking to end with a contradiction?
• Nov 12th 2009, 09:36 PM
redsoxfan325
Yes. Since $L$ is a constant, you can pull it out of the sum.

$\sum_{n=N}^{\infty}\frac{L}{2}=\frac{L}{2}\sum_{n= N}^{\infty}1$

which clearly diverges to $\infty$. This is a contradiction, so $L=0$.
• Nov 13th 2009, 01:18 AM
hjortur
I just wanted to show another way to do this.

Let $\sum_{n=1}^{\infty}x_n=K$
You know that $\sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n = x_N$

So taking the limit:

$\lim_{N\to\infty}x_N=\lim_{N\to\infty}\sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n$

But because you know that the sum is finite:
$=K-K=0$
• Nov 13th 2009, 05:03 AM
HallsofIvy
Yet another way: Let $S_n= \sum_{n=1}^N a_n$ 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 $a_n$ does not go to 0, $S_{n+1}- S_n= a_{n+1}$ does not go to 0. That means that $S_{n}- S_m$ cannot go to 0 as n and m go to 0 independently and so ${S_n}$ is not a Cauchy sequence.

Since every convergent sequence is a Cauchy sequence, ${S_n}$ does not converge and thus $\sum_{n=1}^\infty a_n$ does not converge.