Respuesta :

LammettHash
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.

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