series conv.: sum [n=0, infty] [cbrt(n^2+2)/sqrt(n^3+3)]

[cheffy]

\(\displaystyle \
\sum\limits_{n = 0}^\infty {\frac{{\sqrt[3]{{n^2 + 2}}}}{{\sqrt {n^3 + 3} }}}
\\)

Does this series converge or diverge?

I'm not sure which test I'm supposed to use. Thanks!

 

[pka]

Suppose n>1,
\(\displaystyle \L
\begin{array}{rcl}
\frac{{\sqrt[3]{{n^2 + 2}}}}{{\sqrt {n^3 + 3} }} & > & \frac{{\sqrt[3]{{n^2 }}}}{{\sqrt {n^3 + n^3 } }} \\
& \ge & \frac{{\sqrt[3]{{n^2 }}}}{{\sqrt 2 \sqrt {n^3 } }} \\
& \ge & \frac{1}{{\sqrt 2 \sqrt[6]{{n^5 }}}} \\
\end{array}\)

 

[cheffy]

isn't that asymptotic as 1/(x^(3/2)) and doesn't that converge? meaning that doesn't do anything?

 

[pka]

Because p<1 use the p-series.
\(\displaystyle \L p = \frac{5}{6}\quad \quad \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt 2 \sqrt[6]{{n^5 }}}}}\)

 

[cheffy]

ooh, I misread it. Thanks a lot!