Floor Square Root Summation

Consider the following sum:

$\displaystyle S_k = \sum_{n=1}^{k^2} \frac{1}{2\lfloor \sqrt{n} \rfloor +1} \;\; , k\in N^*$

$\displaystyle S_k = \frac{1}{2k+1} + \sum_{n=1}^{k^2-1} \frac{1}{2\lfloor \sqrt{n} \rfloor +1}$

Let us fix the value of $\lfloor \sqrt{n} \rfloor$ to an integer $m$ . So $m$ can range between $1$ and $k-1$ . (As long as we can see that $n$ is ranging from $1$ to $k^2-1$ ).

The question is : How many possibilities can $\lfloor \sqrt{n} \rfloor$ be equal to $m$ ??

Note that for this to be possible, we have:

$\displaystyle m \leq \sqrt{n} < m+1$

Hence:

$\displaystyle m^2 \leq n \leq (m+1)^2 -1$

Which means that $n$ can take $\big((m+1)^2 -1 - m^2 +1 \big)= (2m+1 )$ values in that range.Therefore we can conclude:

It has $2m+1$ possibilities for $\lfloor \sqrt{n} \rfloor$ to be equal to $m$ .

And Since $m$ can range between $1$ and $k-1$ , We can write our sum as follows:

$\displaystyle S_k = \frac{1}{2k+1} + \sum_{n=1}^{k^2-1} \frac{1}{2\lfloor \sqrt{n} \rfloor +1}$

$\displaystyle = \frac{1}{2k+1} +\sum_{m=1}^{k-1} \frac{1}{2m+1} \times (2m+1)$

$\displaystyle = \frac{1}{2k+1} +k-1$

So , for all non zero natural $k$ , we have:

$\displaystyle \lfloor S_k \rfloor = \lfloor \frac{1}{2k+1} +k-1 \rfloor = \boxed{k-1}$

Now just plug in $k=10$ for this problem.