$\displaystyle \sum_{1}^{\infty}\frac {2^k+k^3}{\sqrt k 3^k}$
how to show if it converges or not?
Ratio test should work here.
sorry i havd a typo ,its
$\displaystyle \sum_{1}^{\infty}\frac {2^k+k^3}{\sqrt k 3^k}$
i tried this law.
i got limit k goes to infinity of
$\displaystyle 2(\frac{k+1}{k})^3 \frac{\sqrt{k}+3^k}{\sqrt{k+1}+3^{k+1}}$
is it ok to solve each fracture as if there where to different limits
and then multiply the results?
I'm not sure what you've done.
You need to evaluate $\displaystyle \displaystyle \lim_{k \to \infty}\left|\frac{a_{k+1}}{a_k}\right|$, which in this case is
$\displaystyle \displaystyle \lim_{k \to \infty}\left|\frac{\frac{2^{k+1}+(k+1)^3}{3^{k+1} \sqrt{k+1}}}{\frac{2^k+k^3}{3^k\sqrt{k}}}\right|$