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Thread: Proof involving series

  1. #1
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by katielaw View Post
    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 $\displaystyle \sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}-\sum_{n=1}^{N}a_n=L-\sum_{n=1}^{N}a_n$. So..
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    Quote Originally Posted by Drexel28 View Post
    Note that $\displaystyle \sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}-\sum_{n=1}^{N}a_n=L-\sum_{n=1}^{N}a_n$. So..
    I don't understand what you're trying to show me..
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by katielaw View Post
    I don't understand what you're trying to show me..
    The idea is that $\displaystyle \sum_{n=N}^{\infty}a_n$ is merely $\displaystyle \sum_{n=1}^{\infty}a_n$ with the first $\displaystyle N$ terms deleted. In other words, $\displaystyle \sum_{n=N}^{\infty}=\sum_{n=1}^{\infty}a_n-\sum_{n=1}^{N}a_n$. So let $\displaystyle \sum_{n=1}^{\infty}a_n=L$, we can say this because the series converges. Thus $\displaystyle \sum_{n=N}^{\infty}a_n=\sum_{n=1}^{\infty}a_n-\sum_{n=1}^{\infty}a_n=L-\sum_{n=1}^{N}$. But then $\displaystyle \lim_{N\to\infty}\left\{\sum_{n=N}^{\infty}a_n\rig ht\}=\lim_{N\to\infty}\left\{L-\sum_{n=1}^{N}a_n\right\}=L-\lim_{N\to\infty}\sum_{n=1}^{N}a_n$...but.
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  5. #5
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    OK, alright, I think I see it now. Thank you.
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