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

When proving convergence the part where $|a_n-a|<\varepsilon$, is it necessary to use a strict inequality? Could I use $\leq$ instead? Thanks.

share|cite|improve this question
HINT If $0 \leq x \leq \epsilon$, then $x$ is strictly less than $2 \epsilon$. – Srivatsan Sep 15 '11 at 14:11
Aha, thank you! – leonard Sep 15 '11 at 14:13
up vote 1 down vote accepted

Welcome to math.stackexchange leonard !

Yes we can use $\leq$ instead, but we usually don't because we set an arbitrary $\epsilon>0$. So if some quantity $M$ has the property that $ M \leq \epsilon$ for any $\epsilon > 0 $, then certainly if we replace the $\epsilon$ by $\epsilon/2 $, we see that we can also say that $ M \leq \epsilon/2 < \epsilon $ for any $\epsilon > 0 $. In the other way, if $M< \epsilon $ for any $\epsilon > 0$, then certainly $ M \leq \epsilon$ as well. Thus, allowing the equality is somewhat redundant and because we would rather be "cleaner" in our definitions, we exclude it.

share|cite|improve this answer
+1 I like the point about arbitrary $\epsilon$. But I want to point out that certain things do change when $\epsilon$ is replaced by $\epsilon/2$. For e.g., as in Levon's answer, the $N(\epsilon)$ threshold in the definition of limits will increase to $N(\epsilon/2)$. However, I agree that this shouldn't be a big deal in most cases. – Srivatsan Sep 15 '11 at 14:20

No, it isn't. If for every $\epsilon$, you can find $N(\epsilon)$ such that $\forall n \ge N(\epsilon), |a_n - a| \le \epsilon$, then just pick an $\epsilon' < \epsilon$ and $N(\epsilon')$ will do the job.

share|cite|improve this answer

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.