Why is $1/n^{1/3}$ convergent? I thought because $p<1$ it would be divergent, but apparently not. Why is that?
 A: Note that $n^{1/3}\to\infty$ as $n\to\infty$. And your sequence is $$n^{-1/3}=\frac{1}{n^{1/3}}$$
ADD It seems you're confusing things. For any $p>0$, $$n^{-p}\to 0$$
However $$\sum_{n\geqslant 1}n^{-p}$$ converges only when $p>1$.
A: The sequence $\{n^{-1/3}\}_{n=1}^\infty$ is convergent, since $n^{-1/3} = \frac{1}{n^{1/3}}$ tends to $0$ as $n$ grows. The series (i.e. sequence of partial sums) resulting in 
$$\sum\limits_{n = 1}^\infty n^{-1/3}$$ 
is divergent, however.
A: You, certainly, know the series $\sum (1/n)$ and know that it is divergent. There is  a nice approach in which we can test the divergence or convergence. That is the Quotient Test or Limit comparison test. According to it, if $$\lim_{n\to\infty}\frac{u_n}{v_n}=A\neq0, ~~\text{or}~~ A=\infty$$ then $\sum u_n$ and $\sum v_n$ have the same destiny. Here, take $\sum v_n=\sum (1/n)$.
A: By the inequality
$$\frac{1}{n}\leq \frac{1}{n^{1/3}}$$
we conclude that the serie $\displaystyle\sum_n \frac{1}{n^{1/3}}$ is divergent .
A: The series $\displaystyle \sum_{n\geqslant 1} n^{-p}$ converges for $p>1$, if you are curious to why, you could search up Cauchy Condensation Test or even the Integral test.  
