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.

`(n^(1/3))/(1-n^(1/3))=\frac1{\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

`(\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))=(n-(n-1))/(n^(2/3)+n^(1/3)(n-1)^(1/3)+(n-1)^(2/3))=\frac1{n^(2/3)+n^(1/3)(n-1)^(1/3)+(n-1)^(2/3)}`

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