<s>[INST] 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:
$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$
 [/INST] 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: $$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}=\sum_{n=2}^{\infty} \frac{1}{n^2-1}=\cdots=\frac{3}{4}$$
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