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Math Help - Prove that if a sum converges, its nth term is 0.

  1. #1
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    Prove that if a sum converges, its nth term is 0.



    I apologize, I'm so new at proofs. This should be very easy, but I'm not sure how to say it. This is what I've come up with. Is it sufficient? Is there another way I should be looking at this?

    Thanks in advance!
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  2. #2
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    Krizalid's Avatar
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    I don't usually prove statements by using contradictions, but here's how I'd solve this:

    Given \sum_{n=1}^\infty a_n<\infty, then \lim_{n\to\infty}\sum_{k=1}^n a_k exists, and so does \lim_{n\to\infty}\sum_{k=1}^{n-1} a_k, hence \underset{n\to \infty }{\mathop{\lim }}\,\left( \sum\limits_{k=1}^{n}{a_{k}}-\sum\limits_{k=1}^{n-1}{a_{k}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( \sum\limits_{k=1}^{n-1}{a_{k}}+a_{n}-\sum\limits_{k=1}^{n-1}{a_{k}} \right)=0, then \underset{n\to \infty }{\mathop{\lim }}\,a_{n}=0, as claimed.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    I apologize, I'm so new at proofs. This should be very easy, but I'm not sure how to say it. This is what I've come up with. Is it sufficient? Is there another way I should be looking at this?

    Thanks in advance!
    Let \sigma_m=\sum_{n=1}^{m}a_n. Since, \lim\text{ }\sigma_m exists (i.e. it converges) we have that it is Cauchy. Thus, for every \varepsilon>0 there exists an N\in\mathbb{N} such that N<\ell\leqslant k\implies\left|\sigma_k-\sigma_\ell\right|=\left|\sum_{n=1}^{\ell}-\sum_{n=1}^{k}\right|=\left|\sum_{n=\ell}^{k}a_n\r  ight|<\varepsilon. Taking \ell=k we see that for every \varepsilon>0 there exists an N\in\mathbb{N} such that N<\ell\implies \left|\sum_{n=\ell}^{\ell}a_n\right|=\left|a_\ell\  right|<\varepsilon. Therefore, a_\ell\to0.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    Let \sigma_m=\sum_{n=1}^{m}a_n. Since, \lim\text{ }\sigma_m exists (i.e. it converges) we have that it is Cauchy. Thus, for every \varepsilon>0 there exists an N\in\mathbb{N} such that N<\ell\leqslant k\implies\left|\sigma_k-\sigma_\ell\right|=\left|\sum_{n=1}^{\ell}-\sum_{n=1}^{k}\right|=\left|\sum_{n=\ell}^{k}a_n\r  ight|<\varepsilon. Taking \ell=k we see that for every \varepsilon>0 there exists an N\in\mathbb{N} such that N<\ell\implies \left|\sum_{n=\ell}^{\ell}a_n\right|=\left|a_\ell\  right|<\varepsilon. Therefore, a_\ell\to0.
    This is what I was trying to say, but I had no idea how to say it. I kept writing things and erasing it because it didn't make sense like I wanted it to

    Thank you so much!
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