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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?2010-09-13
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    Just a plain proof will do :)2010-09-13
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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 http://en.wikipedia.org/wiki/Nth_term_test.2010-09-13

2 Answers 2

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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.2010-09-13
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    What is wrong with the indices?2010-09-13
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    Lona: Robin fixed them, see the edit history to see what was there.2010-09-13
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    This looks like the most clear answer to me, thanks people!2010-09-13
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    Also, "limit n->infty" needs to be added to the RHS of the equation.2010-09-13
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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 0.2010-09-13
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    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.2010-09-13