Clearly, [tex]\dfrac1{k\sqrt{k+2}}<\dfrac1{k\sqrt k}=\dfrac1{k^{3/2}}[/tex] for all [tex]k\ge1[/tex].
Recall that
[tex]\displaystyle\sum_{k\ge1}\frac1{k^p}=\begin{cases}\text{converges}&\text{for }p>1\\\text{diverges}&\text{otherwise}\end{cases}[/tex]
Since [tex]p=\dfrac32>1[/tex], it follows that
[tex]\displaystyle\sum_{k\ge1}\dfrac1{k\sqrt k}[/tex]
also converges, which in turn means that the series
[tex]\displaystyle\sum_{k\ge1}\dfrac1{k\sqrt{k+2}}[/tex]
does too.
