Question

Determine whether the series converge or diverge. If they converge, find the limits.a. an= (n^1/3)/(1-n^1/3)b. an = (n^1/3) - (n^3 -1)^(1/3)

Answer 1

a.

$\dfrac{n^{1/3}}{1-n^{1/3}}=\dfrac1{\frac1{n^{1/3}}-1}$

As $n\to\infty$, the $n^{-1/3}$ term will converge to 0, so $a_n\to-1$.

b. If you mean

$n^{1/3}-(n^3-1)^{1/3}$

then the sequence diverges, since $(n^3-1)^{1/3}$ behaves like $n$, and $n>n^{1/3}$ for $n>1$.

But if you mean

$n^{1/3}-(n-1)^{1/3}$

rewrite as

$\dfrac{\left(n^{1/3}-(n-1)^{1/3}\right)\left(n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}\right)}{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}=\dfrac{n-(n-1)}{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}=\dfrac1{n^{2/3}+n^{1/3}(n-1)^{1/3}+(n-1)^{2/3}}$

which converges to 0 as $n\to\infty$.