There's no hard-and-fast method for determining which test to use; your best bet is to use whichever test best fits the situation, i.e. the one whose hypotheses are information you already know about the function or can easily prove about it. If no such test meets this condition, it can become a matter of trial and error to determine which one works. Beyond that, you just need practice.
Suppose you have a series $\sum a_i$ and you want to prove its convergence or divergence. Let's see when each test would be appropriate:
Limit comparison and direct comparison most effective if the terms in your sequence resemble a familiar series, such as a $p$-series. For example, suppose $$a_i = \frac{1}{i^3 + 2i}$$ which looks suspiciously looks like a convergent $p$-series (where $p = 3$). Indeed, $$\lim_{i \to \infty} \frac{i^3 + 2i}{i^3} = 1 \in (0, \infty)$$ so $a_i$ converges. Limit comparison is especially good for verifying series that intuitively seem like they should converge or diverge but it's hard to prove directly.
If $a_i$ looks like a function $f(i)$ whose integral you are comfortable computing, you should use the integral test. For example, suppose you didn't know the $p$-series test; then it would be easy to check that $a_i = i^{-p}$ converges if and only if $p > 1$simply by computing the integral $$\int_0^{\infty} x^{-p} ~dx$$ which isn't particularly hard. Be especially on the lookout for functions which look like they could be integrated by parts or $u$-substitution.
The ratio and root tests are more situational. The root test is mostly useful if $a_i$ is of the form $a_i = b_i^i$ for some sequence $b_i$ which you can easily compute the limit of, because then $a_i^{1/i} = b_i$ and one only has to compute $\limsup b_i$. Of course, if $\limsup b_i = 1$ then this test is inconclusive and this test is a red herring. Similarly, the ratio test is good for series whose terms can be easily cancelled by multiplication and division, but could be a red herring.
Best Answer
$\lim\limits_{k\rightarrow +\infty}k^3e^{-\sqrt{k}}=0$ and since $\sum\frac{1}{k^2}$ converges, $\sum ke^{-\sqrt{k}}$ converges aswell.