Terminology.
The sequence converges:
lim[n→∞] n/√(n^3 + 2) = 0
TychaBrahe covers this, although she calls this a series.
The series diverges:
∑[n=1,∞] n/√(n^3 + 2) = ∞
Barney Good demonstrates this quite thoroughly.
Because you didn't include a summation sign in your question, it could be taken either way, if you erred in calling a sequence a series.
BTW, Barney, the nth term test *can* show convergence, but only for an alternating series, which, of course, this one is not. So you're right that in this case, that test fails to decide the question.
* * * CAUTION: The claim that this test can show convergence for an alternating series, turns out to be false, as developed below. * * *
EDIT:
@ B.G.: OK, I get that you're not referring to that, but correct me if I'm wrong, say, with a counterexample. If a sequence converges to 0, and if beyond some finite point, its signs alternate, then the series converges. This is called conditional convergence if the absolute series diverges, but it is convergence, nonetheless. That is, the limit of the sequence of partial sums converges, and no further tests are needed in that case.
BTW, BG, I want to give a shout out to the clowns who've given you 3 TD's, because your answer is absolutely solid, and instructive. Shame on those clowns!
EDIT2:
BG: An infinite sequence can be any infinite list of numbers, and can do wild things before settling down to a pattern, or it can just never settle into a pattern. You could take a well-behaved I.S. and append n terms of any kind to the beginning, and you'd have a new I.S. You could also take that well-behaved I.S. and stick funny terms into it every so often, forever, and that would be another legitimate I.S. So when arriving at rules for finding the behavior of a series, your rules have to be able to handle anything that might come along. Your condition 2 is meant to do that -- I believe you meant "the absolute sequence (that is, the sequence of absolute values) has to be monotonic decreasing." So if there's a way to violate condition 2 while still satisfying condition 1, then that should form the basis for a counterexample to what I said. It's been too long since my course in Infinite Series, so I'll have to do some reviewing.
But as for settling down only after some finite point -- the tests for overall behavior of an infinite sequence or series still apply, as long as you can 'quarantine' a finite number of unruly terms in such a way that the remaining ones satisfy the test. That is sufficient to draw the conclusion from the test.
Also, I see that you could strengthen your main answer by revising the first comparison in your chain:
n/√(n^3 + 2) ≥ n/(3n^3) = 1/(3√n) > 1/3n --> diverges.
[Actually, I see that nyc_kid has done this.]
EDIT3:
BG: I see what can 'cook' my statement for convergence for an alternating series, by way of a counterexample. If you let all the odd terms be
a[n] = 1/n; n = 1, 3, 5, ...
and all the even terms be
a[n] = -1/2^n; n = 2, 4, 6, ...
so overall it goes:
1 - 1/4 + 1/3 - 1/16 + 1/5 - 1/64 + - ...
then
∑[j=1,∞] a[n] = ∑[j=1,∞] 1/(2j - 1) - ∑[j=1,∞] 1/4^j
and the second sum converges to 1/3, while the first sum diverges (by comparison with (1/2)*(1/n), for n=1,2,3...; 1 > 1/2, 1/3 > 1/4, 1/5 > 1/6, ...), making the overall sum diverge (∞ - 1/3 = ∞). The Leibniz test's 2nd requirement ("L2"), namely, that the absolute sequence be monotonic decreasing (forever starting from some finite point), smokes this out and rejects it. That is, L2 fails to decide [con/di]vergence. (Since L2 is a sufficient, but not necessary condition for convergence, because we could have made the odd terms = 1/n^2, and it would still fail L2, but would converge to:
(π^2)/8 - 1/3 )