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I've been using the sentence:

If a series converges then the limit of the sequence is zero

as a criterion to prove that a series diverges (when $\lim \neq 0$) and I can understand the rationale behind it, but I can't find a formal proof.

Can you help me?

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How formal do you want the proof? As in a formal proof system such as Mizar or Coq? –  Robin Chapman Sep 13 '10 at 10:57
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Just a plain proof will do :) –  Lona Payne Sep 13 '10 at 11:09
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I find it a little strange that you had trouble finding a formal proof. For instance, every calculus textbook I have ever seen has a proof, as do many elementary analysis textbooks. Also see en.wikipedia.org/wiki/Nth_term_test. –  Pete L. Clark Sep 13 '10 at 17:34
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2 Answers 2

up vote 10 down vote accepted

Yes.

$$\lim_{n \to \infty} \left ( \sum_{k = 1}^{n + 1} a_k - \sum_{k = 1}^{n} a_k \right ) = \lim_{n \to \infty} a_{n + 1} $$ And both sums will converge to the same number so the limit is zero. This is by far the easiest proof I know.

This is the Cauchy criterion in disguise by the way, so you could use that too.

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You might want to correct your indices in the summations. –  Asaf Karagila Sep 13 '10 at 10:59
    
What is wrong with the indices? –  Lona Payne Sep 13 '10 at 11:15
    
Lona: Robin fixed them, see the edit history to see what was there. –  J. M. Sep 13 '10 at 11:18
    
This looks like the most clear answer to me, thanks people! –  Lona Payne Sep 13 '10 at 11:32
    
Also, "limit n->infty" needs to be added to the RHS of the equation. –  Douglas S. Stones Sep 13 '10 at 12:04
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If we know that the sequence converges and merely wish to show it converges to zero, then a proof by contradiction gives a little more intuition here (although the direct proofs are simple and beautiful). Assume $a_n\to a$ with $a>0$, then for all $n>N$ for some large enough $N$ we have $a_n > a/2$ (take $\varepsilon = a/2$ in the definition of the limit). Now the sum diverges: $\sum_{n>N}a_n > \sum_{n>N}a/2 = \infty$. A similar argument works when $a<0$.

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You are mistakingly assuming that the sequence actually converges to something. The other proofs also show that the limit exist, AND it is 0. –  Andrea Ferretti Sep 13 '10 at 11:24
    
Indeed, I thought that this was the question - given that the sequence converges, then the limit is zero and not something else. I'll add this to my answer. –  Gadi A Sep 13 '10 at 11:52
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