I was trying to prove the convergence of the series $$ \sum_{n=1}^{\infty} \frac{n^{\frac{1}{3}}}{n^{\frac{5}{3}} -3} $$ My original approach was to try and bound the sequence $a_n = \frac{n^{\frac{1}{3}}}{n^{\frac{5}{3}} -3}$ by some sequence $b_n = \frac{1}{n^k}$ for some $k >1$ and for all $n > N$ (given some fixed natural number $N$). The idea here is that the series of such $b_n$'s would converge by the p-series test, and hence, by the comparison test we could conclude that our original series converges. The problem I encountered is that bounding the series in such a way gives very little breathing room for the possible values of $k$. In fact, for as soon as $k = 1.4$ the inequality $$ \frac{n^{\frac{1}{3}}}{n^{\frac{5}{3}} -3} < \frac{1}{n^{1.4}} $$ is no longer true for sufficiently large $n$. My solution was to bound the sequence $a_n$ by a sum of p-series instead of just one p-series. I managed to prove that $$ \frac{n^{\frac{1}{3}}}{n^{\frac{5}{3}}-3} < \frac{1}{n^{\frac{4}{3}}} + \frac{1}{n^{\frac{5}{3}}} \qquad \forall n \ge 4 $$ which is enough to prove convergence using the comparison test since $\zeta\left(\frac{4}{3}\right) + \zeta\left(\frac{5}{3}\right) < \infty$. However, even though this works, proving the above inequality is not as simple as I would like, but I couldn't think of an easier way to prove the convergence.
Can anyone think of an easier (or more direct) way to prove the convergence of the series in question? Any convergence test works, it doesn't necessarily have to be by direct comparison. Any and all thoughts are welcome. Thank you!
Usually, the short way uses asymptotic equivalence, which requires the general term of the series to be ultimately positive, which is the case here. Remember equivalence is compatible with multiplication and division (but not with addition nor subtraction):