How could I evaluate $\sum_{r=1}^\infty{\frac{1}{r\sqrt{r+1}}}\;?$ How could I evaluate $\displaystyle\,\sum_{r=1}^\infty{\frac{1}{r\sqrt{r+1}}}\;?$
I have tried the substitution $r=k^2-1$, but this yields a sum with a square root as the starting point ($\sqrt2$).
$$\int_1^\infty{\frac{1}{\lfloor x^2 \rfloor}}=\int_1^\sqrt{2}{1}dx+\int_\sqrt{2}^\sqrt{3}{\frac{1}{2}}dx+\int_\sqrt{3}^\sqrt{4}{\frac{1}{3}}dx\,+...$$ I think.

Considering $$\sum_{r=1}^\infty{\frac{1}{r\sqrt{r+1}}}$$ let us use
$$\frac{1}{r\sqrt{r+1}}=\sum_{n=0}^\infty (-1)^n \frac{1}{4^n} \binom{2 n}{n} r^{-n-\frac{1}{2}}$$
$$\sum_{r=1}^\infty{\frac{1}{r\sqrt{r+1}}}=\sum_{n=0}^\infty (-1)^n \frac{1}{4^n} \binom{2 n}{n}\, \zeta \left(\frac{2 n+3}{2} \right)$$ but this is rather slow even if alternetive.
If you use Euler-McLaurin summation formula (limited to $f^{(5)}(r)$), you have
$$\sum_{r=1}^\infty{\frac{1}{r\sqrt{r+1}}}\sim \frac{6174871}{10321920 \sqrt{2}}+2 \sinh ^{-1}(1)=2.1857\cdots$$  instead of $2.1840\cdots$
