Take the 2-minute tour ×
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.

$\displaystyle\sum_{n=0}^{\infty}\sqrt{n}(\sqrt{n^4+1}-n^2)$

Ok I was really hesitant to post this because it's such an elementary question, and I really should be able to do this. However, I've been looking at this for over half an hour and I've tried many methods, but I'm not able to get the series into a nice form where I can take the limit. Any ideas on how to tackle this problem?

share|improve this question
5  
Have you tried rationalizing? Note that n^2 = sqrt{n^4}. –  Qiaochu Yuan Dec 8 '10 at 0:28

1 Answer 1

up vote 9 down vote accepted

Just note that $$\sqrt{n^4 + 1} - n^2 = \frac{( \sqrt{n^4 + 1} - n^2 )( \sqrt{n^4 + 1} + n^2 )}{\sqrt{n^4 + 1} + n^2} = \frac{n^4 + 1 - n^4}{\sqrt{n^4 + 1} + n^2} = \frac{1}{\sqrt{n^4 + 1} + n^2}$$ then you can use the comparison test.

share|improve this answer
    
How'd you get that? –  maq Dec 8 '10 at 0:35
    
@f-Prime I edited my answer to show what I did. As Qiaochu commented, this is just a rationalization. –  Adrián Barquero Dec 8 '10 at 0:39
    
@f-Prime: recall that a^2 - b^2 = (a - b)(a + b). In this case a = sqrt{n^4+1}, b = n^2, a^2 - b^2 = 1, and we are trying to understand the behavior of a - b. But this is just 1/(a + b) and it is easy to see what the behavior of a + b is. This is just one of those tricks you learn to look out for. –  Qiaochu Yuan Dec 8 '10 at 0:41
    
Very nice, thanks! –  maq Dec 8 '10 at 0:42

Your Answer

 
discard

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.