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

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 $n=0$. 
