1. Proof involving series

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.

2. 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..

3. 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..

4. 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.

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