TITLE: Computing $ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$
QUESTION [7 upvotes]: I would like to prove the existence and the exact value of the following series:
$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$

REPLY [5 votes]: 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
$$s={1\over 2}\sum_{r=1}^\infty \Bigl({1\over r}-{1\over r+2}\Bigr)={1\over2}\bigl(1+{1\over2}\Bigr)={3\over4}\ .$$