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I tried to solve the following problem.

What is the integer part of $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{(2n+1)^2}}=\sum_{k=0}^{2(n^2+n)} \frac 1{\sqrt{2k+1}} ?$$

I tried using some inequalities( by grouping 1,3,5,7 / 9,11,13,...,25/ ), but I failed.

How can I compute the integer part of this sum?

  • 0
    Why do you write $\,\sqrt{(2n+1)^2}\,$? This equals $\,2n+1\,$ , which is not what you wrote in the denominators...2012-11-17
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    @DonAntonio I suppose that it is meant to be like this, i.e. there are $\frac{(2n+1)^2+1}{2}$ terms in that sum.2012-11-17
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    I meant $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{(2n+1)^2 - 2}} + \frac{1}{\sqrt{(2n+1)^2}}$. Maybe we cannot compute this for any case??2012-11-17
  • 0
    Have you tried approximating your sum by an integral?2012-11-17

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