The first series is positive, so it either converges absolute, or diverges. My gut reaction was to try some back-of-the-envelope comparison.
Let's see: $\sum \frac{1}{e^{k}}$ converges, and $\frac{1}{e^{k^3}}\leq \frac{1}{e^k}$, so $\sum\frac{1}{e^{k^3}}$ converges. The exponential dominates $\sqrt{k}$, so personally, I would expect the series to converge. I would try an integral test, but it looks like an annoying function to integrate. So perhaps we can find a series $\sum\frac{p(k)}{e^{k^3}}$ for some function $p(k)$, such that $\int_1^{\infty}p(x)e^{-x^3}\,dx$ is easy to integrate.
This is doable, but as user6312 pointed out, the Ratio Test will do the job as well.
The second series is (eventually) alternating. Try the Alternating Series Test. To check the absolute convergence, try an Integral test. You'll need a change of variable. Hint. If you set $u=\ln x$, what happens?