Proof involving series

**katielaw** · December 1st 2009, 04:41 PM

Show that if ∑a_n converges, then ∑ from N to ∞ of a_n -> 0 as N goes to ∞.

My first thought was to basically let a_n = s_n - s_(n-1). And since a_n and s_n are both convergent, just take the limits of s_n and s_(n-1) as n -> ∞ and show they both equal zero.

**Drexel28** · December 1st 2009, 04:52 PM

Originally Posted by katielaw

Show that if ∑a_n converges, then ∑ from N to ∞ of a_n -> 0 as N goes to ∞.

My first thought was to basically let a_n = s_n - s_(n-1). And since a_n and s_n are both convergent, just take the limits of s_n and s_(n-1) as n -> ∞ and show they both equal zero.

Note that $\sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}-\sum_{n=1}^{N}a_n=L-\sum_{n=1}^{N}a_n$ . So..

**katielaw** · December 1st 2009, 05:15 PM

Originally Posted by Drexel28

Note that $\sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}-\sum_{n=1}^{N}a_n=L-\sum_{n=1}^{N}a_n$ . So..

I don't understand what you're trying to show me..

**Drexel28** · December 1st 2009, 05:23 PM

Originally Posted by katielaw

I don't understand what you're trying to show me..

The idea is that $\sum_{n=N}^{\infty}a_n$ is merely $\sum_{n=1}^{\infty}a_n$ with the first $N$ terms deleted. In other words, $\sum_{n=N}^{\infty}=\sum_{n=1}^{\infty}a_n-\sum_{n=1}^{N}a_n$ . So let $\sum_{n=1}^{\infty}a_n=L$ , we can say this because the series converges. Thus $\sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}a_n-\sum_{n=1}^{\infty}a_n=L-\sum_{n=1}^{N}$ . But then $\lim_{N\to\infty}\left\{\sum_{n=N}^{\infty}a_n\rig ht\}=\lim_{N\to\infty}\left\{L-\sum_{n=1}^{N}a_n\right\}=L-\lim_{N\to\infty}\sum_{n=1}^{N}a_n$ ...but.

**katielaw** · December 1st 2009, 05:28 PM

OK, alright, I think I see it now. Thank you.

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