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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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    Have you tried approximating your sum by an integral?2012-11-17

2 Answers 2

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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....2013-12-29
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Hint: The sum is an approximation for $c\int_a^b \frac1{\sqrt x} \, dx $ with suitable chocies of $a,b,c$.