I'm assuming the sum in question is ∑ sin[(6n^{2} + 5)/(n^{5} + 5)].

I prefer the limit comparison test for series such as this.

Note

lim _{n→∞} sin[(6n^{2} + 5)/(n^{5} + 5)] / [(6n^{2} + 5)/(n^{5} + 5)]

= lim _{x→0} sin(x)/x

= 1.

Since the limit is a positive finite number, the given series and the series ∑ [(6n^{2} + 5)/(n^{5} + 5)] either both converge or both diverge.

Next, note

lim _{n→∞} [(6n^{2} + 5)/(n^{5} + 5)] / (1/n^{3})

= lim_{ n→∞} [(6n^{5} + 5n^{3})/(n^{5} + 5)]

= 6.

Since the limit is a positive finite number, the series ∑ [(6n^{2} + 5)/(n^{5} + 5)] and the series ∑ (1/n^{3}) either both converge or both diverge. But of course ∑ (1/n^{3}) converges. So the given series converges.

Now, we should also check that the sequences (1/n^{3}) and [(6n^{2} + 5)/(n^{5} + 5)] and sin [(6n^{2} + 5)/(n^{5} + 5)] have only positive terms, since this is a condition of the limit comparison test. This is obvious in the first two cases for all n ≥ 1.

To see that sin [(6n^{2} + 5)/(n^{5} + 5)] is positive for all n ≥ 2, put f(x) = (6x^{2} + 5)/(x^{5} + 5). Then

f'(x) = [x(-18x^{5} - 25x^{3} + 60)] / (x^{5} + 5)^{2}.

For x ≥ 2, the sign of f' is clearly negative while the sign of f is clearly positive. Together with the fact that f(2) = 29/37 < π, it follows that 0 < f(x) < π for all x ≥ 2. Thus, for n ≥ 2, we must have sin[ f(n) ] > 0.