1. ## convergence question..

$\displaystyle \sum_{1}^{\infty}\frac {2^k+k^3}{\sqrt k 3^k}$

how to show if it converges or not?

2. ## Re: convergence question..

Ratio test should work here.

3. ## Re: convergence question..

Have you tried the ratio test?

4. ## Re: convergence question..

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?

5. ## Re: convergence question..

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

$\displaystyle \displaystyle \lim_{k \to \infty}\left|\frac{\frac{2^{k+1}+(k+1)^3}{3^{k+1} \sqrt{k+1}}}{\frac{2^k+k^3}{3^k\sqrt{k}}}\right|$

6. ## Re: convergence question..

thats was done but we get a multiplication
of two differenent fractures
which could be solve apart only