For [tex]k\ge1[/tex], you have [tex]\sin\dfrac\pi k\le1[/tex], and so
[tex]\dfrac1{k^5}\sin\dfrac\pi k\le\dfrac1{k^5}[/tex]
If you're familiar with [tex]p[/tex]-series, you know that
[tex]\displaystyle\sum_{k\ge1}\frac1{k^p}[/tex]
converges for [tex]p>1[/tex]. Here we have [tex]p=5[/tex]. So the given series converges by comparison, and since it converges, it must converge absolutely.
