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I have the following infinite series:

$$\sum_{n=1}^{\infty}{\left(1-\frac{1}{2n}\right)^{n^2}}.$$

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

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4 Answers

up vote 10 down vote accepted

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

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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
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Hint: You may want to try the ratio test.

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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
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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.

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Do you want another idea? Good:

$$\left(1-\frac{1}{2n}\right)^{n^2}=\left[\left(1-\frac{1}{2n}\right)^{2n}\right]^{n/2}\le\left(e^{-1}+0.1\right)^{n/2}$$

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.

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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
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