Does $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{} k+1}}$ converge? Does the following series converge? $$\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{|} k+1}}$$ 
I tried using the ratio test and the comparison test but I wasn't able to solve this. 
I think I should try manipulating the denominator to use comparison test but I can't figure out how?
 A: Note that $\frac{1}{k\sqrt{k+1}} < \frac{1}{k \sqrt{k}} = k^{-1.5}$.
Hence, by the comparison test, $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}} < \displaystyle\sum_{k=1}^n k^{-1.5} < \infty$ for all $n$. Hence $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}}$ converges, and in fact it is somewhere  close to $2.04288$ by Wolfram Alpha.
A: A slight variation of robjonh's fine answer through creative telescoping.
We may check in advance that for every $k\geq 1$ the inequality
$$ \frac{1}{k\sqrt{k+1}}\leq \frac{2}{\sqrt{k-\frac{1}{5}}}-\frac{2}{\sqrt{k+\frac{4}{5}}}\tag{1}$$
holds, hence it follows that:
$$ \sum_{k\geq 1}\frac{1}{k\sqrt{k+1}}\leq \frac{2}{\sqrt{1-\frac{1}{5}}}=\color{red}{\sqrt{5}}. \tag{2}$$
A: Taking the terms in groups of $2^n$,
$$\frac1{\sqrt2}<1,\\
\frac1{2\sqrt3}+\frac1{3\sqrt4}<\frac1{\sqrt2},\\
\frac1{4\sqrt5}+\frac1{5\sqrt6}+\frac1{6\sqrt7}+\frac2{7\sqrt8}<\frac1{\sqrt{2^2}},\\
\frac1{8\sqrt9}+\frac1{9\sqrt{10}}+\frac1{10\sqrt{11}}+\frac2{11\sqrt{12}}+\frac1{12\sqrt{13}}+\frac1{13\sqrt{14}}+\frac1{14\sqrt{15}}+\frac2{15\sqrt{16}}<\frac1{\sqrt{2^3}},\\\cdots
$$
and the sum is bounded by a geometriec series of common factor $1/\sqrt2$.

The property will remain true replacing the square root by any positive power. (And by a similar lower bound you will show divergence for any non-positive power.)
A: For $k\ge1$, $\sqrt{k+1}\le\sqrt{2k}$. Therefore,
$$
\begin{align}
\frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}
&=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}\\
&=\frac1{\sqrt{k}\sqrt{k+1}\left(\sqrt{k+1}+\sqrt{k}\right)}\\
&\ge\frac1{\left(1+\sqrt2\right)k\sqrt{k+1}}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{k\sqrt{k+1}}
&\le\left(1+\sqrt2\right)\sum_{k=1}^\infty\left(\frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}\right)\\
&=1+\sqrt2
\end{align}
$$
So the series converges.

Using $8$ terms of the Euler-Maclaurin Sum Formula applied to $14$ terms of the Taylor Series for $\frac1{k\sqrt{k+1}}$ and comparing to $1000$ terms of the actual sum gives
$$
\sum_{k}\frac1{k\sqrt{k+1}}=2.18400947026785195289473415785294907
$$
A: Using the fact that square root of x+1 > square root of (x), for x>0 and p test, it converges
