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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?

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Why do you write $\,\sqrt{(2n+1)^2}\,$? This equals $\,2n+1\,$ , which is not what you wrote in the denominators... –  DonAntonio Nov 17 '12 at 14:00
    
@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. –  Dan Shved Nov 17 '12 at 14:07
    
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?? –  Mahtsolver11 Nov 17 '12 at 14:08
    
Have you tried approximating your sum by an integral? –  nayrb Nov 17 '12 at 14:15

2 Answers 2

A more general sum can be bounded as follows:

$$ \int_0^{n+1} \frac{1}{\sqrt{2 x + 1}} dx \leq \sum_{k=0}^n \frac{1}{\sqrt{2 k + 1}} \leq 1 + \int_0^n \frac{1}{\sqrt{2 x + 1}}dx $$

or, computing the integrals

$$ \sqrt{2 n + 3} -1 \leq \sum_{k=0}^n\frac{1}{\sqrt{2 k + 1}} \leq \sqrt{2 n + 1}. $$

The second inequality is strict if $n > 0$. If $2n + 1$ is a square, say $m^2$ then

$$ m^2 < 2n+3 = m^2 + 2 < (m+1)^2 $$

and so the integral part of $\sqrt{2n+3}$ is $m$. From this it follows that the integral part of the sum is $m - 1$ if $n > 0$ and $1$ if $n=0$.

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nice solution WimC.... –  juantheron Dec 29 '13 at 12:25

Hint: The sum is an approximation for $$c\int_a^b \frac1{\sqrt x} \, dx $$ with suitable chocies of $a,b,c$.

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