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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?

learner
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Lona Payne
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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 http://en.wikipedia.org/wiki/Nth_term_test. – Pete L. Clark Sep 13 '10 at 17:34

2 Answers2

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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.

JT_NL
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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 $a<0$.

Gadi A
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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