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

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- Nov 12th 2009, 08:25 PMamm345Limit 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 xn=0 - Nov 12th 2009, 09:07 PMredsoxfan325
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... - Nov 12th 2009, 09:34 PMamm345
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 PMredsoxfan325
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 $\displaystyle L=0$. - Nov 13th 2009, 01:18 AMhjortur
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:

$\displaystyle =K-K=0$ - Nov 13th 2009, 05:03 AMHallsofIvy
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.