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

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?

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    $\begingroup$ Have you tried rationalizing? Note that n^2 = sqrt{n^4}. $\endgroup$ – Qiaochu Yuan Dec 8 '10 at 0:28
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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.

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  • $\begingroup$ How'd you get that? $\endgroup$ – Snowman Dec 8 '10 at 0:35
  • $\begingroup$ @f-Prime I edited my answer to show what I did. As Qiaochu commented, this is just a rationalization. $\endgroup$ – Adrián Barquero Dec 8 '10 at 0:39
  • $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Dec 8 '10 at 0:41

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