Since lim n→ infinity (.............) = ...........

Notice that
[tex]\dfrac{4n+1}{8n+2}=\dfrac{4n+1}{2(4n+1)}=\dfrac12[/tex]
So by the root test,
[tex]\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac12\right)^{2n}\right|} = \frac14 < 1[/tex]
and so the series converges (absolutely).