Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to know what are common methods can be used to show that an infinite sequence converges. From what I know so far,

  1. If a sequence is bounded and monotonic increasing/decreasing then it converges.
  2. Using definition of limit.
  3. Another method that I saw online, is to assume the sequence approaches a limit $L$, then solve for $L$, but I'm not totally convinced that this approach is correct. For example, the Fibonacci ratio sequence, to prove the limit of $$\displaystyle\lim_{n\to\infty} \dfrac{a_{n+1}}{a_n}$$ exists, they claim that: $$1 + \dfrac{1}{L} = L$$ proof for 3

So I wonder could anyone could share me some most commonly used method for proving the limit of an infinite sequence exists that I'm not aware of? Any suggestion or ideas would be appreciated.

share|cite|improve this question
Number $3$ is a method of finding a limit given that it exists. It doesn't prove actual convergence. It merely says that if it converges, then it must converge to this. – EuYu Oct 15 '12 at 1:18
@EuYu: I see. Thanks. – Chan Oct 15 '12 at 1:20
up vote 1 down vote accepted

One of the most powerful tools in calculus(but not only):

Cauchy's criterion for convergence

share|cite|improve this answer

Here is a theorem tells you if a sequence converges to zero or diverges to infinity,

Theorem: If ${a_n}$ be a sequence such that $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}= a\,,$ then

1) if $|a|<1$, then $\lim_{n\to \infty}a_n =0 \,,$

2) if $ a>1$, then $\lim_{n\to \infty}|a_n| =\infty \,.$

See here for applications.

share|cite|improve this answer
If $a_n=-2^n$, then $\frac{a_{n+1}}{a_n}=2$. – robjohn Oct 20 '12 at 7:11
@robjohn: Thanks for the comment. – Mhenni Benghorbal Jan 21 '13 at 19:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.