Computing $ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$ I would like to prove the existence and the exact value of the following series:
$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$
 A: As all terms of the given series $s=\sum_{n=1}^\infty a_n$ are $\geq0$ we may collect them in packets and write
$$s=\sum_{r=1}^\infty\left(\sum\nolimits_{r^2\leq n<(r+1)^2} a_n\right)\ .$$
Note that in the inner sum only the last term, corresponding to $n=(r+1)^2-1$, is nonzero and has the value $${1\over(r+1)^2-1}={1\over (r+2) r}={1\over2}\Bigl({1\over r}-{1\over r+2}\Bigr)\ .$$
It follows that the outer sum is a telescoping series, and we obtain
$$s={1\over 2}\sum_{r=1}^\infty \Bigl({1\over r}-{1\over r+2}\Bigr)={1\over2}\bigl(1+{1\over2}\Bigr)={3\over4}\ .$$
A: Work on Thomas Andrews's hint (That is a very good hint)
Let me give a following hint
$$ \left\lfloor \sqrt{(3+1)}  \right\rfloor = 2,   \hspace{3pt} \left\lfloor \sqrt{(2+1)}  \right\rfloor = 1, \text{ why?} $$
$$ \left\lfloor \sqrt{(8+1)}  \right\rfloor = 3,  \hspace{3pt} \left\lfloor \sqrt{(7+1)}  \right\rfloor = 2, \text{ why?} $$
Try working towards these observed values, and in general for what values are they not equal?
The answer is $$\frac{3}{4}$$ 
A: If $ n+1 $ is a square,  $\lfloor \sqrt{n+1} \rfloor=\sqrt{n+1} $
$$ 0<\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<1$$
So: $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$
So: $$ \lfloor \sqrt{n} \rfloor= \sqrt{n+1}-1$$
$$ \lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor=1 $$
I have proved:
$n+1$ is a square $\Longrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$  
Now I must show that 
$ \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor \Longrightarrow $ $n+1$ is a square
If $ n+1$ is not a square:
If $ n $ is a square, $\lfloor \sqrt{n} \rfloor=\sqrt{n}$.
 As
 $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$
$ \lfloor \sqrt{n+1} \rfloor= \sqrt{n} $ 
So: $  \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $ 
If $ n $ is not a square, there exists $a\in \mathbb{N} $ such that 
$$ a^2<n<n+1<(a+1)^2$$
So:
$$ a<\sqrt{n}<\sqrt{n+1}<a+1$$
So : 
$$  \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $$
Finally:
$n+1$ is a square $\Longleftrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$  
So: $$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}=\sum_{n=2}^{\infty} \frac{1}{n^2-1}=\cdots=\frac{3}{4}$$
