In this mathematical text they are giving a proof for the first statement using a proof by contradiction that I perfectly understand, but I just don't understand why they did not just give this counterexample: $a_n = \frac{1}{n} : n\in\mathbb{N}$

This sequence has the property that all of its elements are greater than zero, and it converges to 0. Since 0 is not less than zero, we get a contradiction.

  • 1
    $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. $\endgroup$
    – 5xum
    Commented Dec 6, 2021 at 12:24

2 Answers 2


Your counterexample is not a counterexample. The statement in question says the following:

Let $a_n$ be a convergent sequence and $\lim_{n\to\infty}a_n=a$. Then, if $a_n\geq 0$ for all $n$, then $a\geq 0$.

Your example merely shows that there exists some sequence $a_n$ such that all its values are positive, and the limit is $0$. The existence of one such sequence does nothing to prove the original statement. It is also not in conflict with the original statement, but it does not prove the original statement.

Also, note that the mere existence of any single sequence can never be proof enough of the original statement. This is because the original statement is talking about all sequences with a given property.

  • $\begingroup$ If I understand right, my counterexample would be a counterexample if the negation of the original statement were " for all the sequences with the given property we have a < 0". Right? $\endgroup$
    – Denis
    Commented Dec 6, 2021 at 12:38
  • $\begingroup$ Yes. That's right. Your counterexample is just what you need to show the converse of the statement is false. $\endgroup$ Commented Dec 6, 2021 at 12:40
  • $\begingroup$ @ShinobiSan Correct. $\endgroup$
    – 5xum
    Commented Dec 6, 2021 at 12:48

$a_n >0 ,\space \space \forall n>N $ and $a_n \to a $ $\implies a\ge 0$

Proof: Suppose, $a<0$

Then, $a< \frac{a}{2} <0$

Given, $a_n \to a$.

Then the interval $(\frac{3a}{2},\frac{a}{2})$ contains all but finitely many terms of $(a_n).$

This contradict the hypothesis that $a_n>0 \space \space \forall n>N$ .

In your case, $\frac{1}{n} >0$

And, $\lim (\frac{1}{n})=0\ge 0$

So, it is an example and doesn't contradict our statment.


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