Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 have the following infinite series:


How can I check whether this infinite series is convergent or diverging?

share|cite|improve this question
up vote 9 down vote accepted

HINT The $n^{th}$ term goes as $\sim e^{-n/2}$.

share|cite|improve this answer
I kept staring trying to figure out how to make it look like $\left(1-\frac1n\right)^n$... +1 – oldrinb Apr 21 '13 at 18:08

Hint: You may want to try the ratio test.

share|cite|improve this answer
Tried, and i've got L = 1. – Billie Apr 21 '13 at 18:58
@user1798362 : the true value is $L=1/\sqrt{e}$ as you can see here, so the sum converges. However, if you have trouble deriving this limit value, just find a sequence that is "greater than" this one and easier to evaluate, and has a limiting value less than or equal to 1 (thats what I did). – Coffee_Table Apr 21 '13 at 19:36

Hint. You may find it more pleasurable to apply the root test, which is essentially equivalent to using user17762's asymptote or Coffee_Table's methodology.

share|cite|improve this answer

Do you want another idea? Good:


Since the rightmost expression is one belonging to a geometric sequence with common quotient less than one its series converges and thus ours does as well.

share|cite|improve this answer
In your last expression, couldn't you just have $e^{-1}$? Since as you used here, $$\left (1-\frac{1}{2n} \right)^{2n} \overset {n \rightarrow \infty}{\rightarrow} e^{-1}.$$ – Coffee_Table Apr 22 '13 at 12:39
@Coffee_Table, For that I would have to prove the sequence is monotone increasing, which I don't think is true but anyway I don't want to mess with that. I just use the limit of the inner part is $\,e^{-1}\,$ so for all $\,n\,$ big enough the inequality is true... – DonAntonio Apr 22 '13 at 12:42
I think I didn't clarify what I meant; I mean why couldn't you have $(e^{-1})^{n/2}$ instead of $(e^{-1}+0.1)^{n/2}$? – Coffee_Table Apr 22 '13 at 12:47
You did clarify, @Coffee_Table : I wanted to propose using the comparison test. For the inequality above to be "obviously" true the right hand side must be something clearly greater than or equal the LHS. The number $\,e^{-1}\,$ doesn't make the cut as it is not true that $\,\left(1-\frac{1}{2n}\right)^{2n}\,$ is monotone increasing... It though has $\,e^{-1}\,$ as limit, which already makes the given inequality true for all $\,n>M\,$ , for some $\,M\in\Bbb N\,$... – DonAntonio Apr 22 '13 at 13:15
Ah, I see. Thanks for the explanation. – Coffee_Table Apr 22 '13 at 16:33

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.