Question

Does `a_n = (n+1)^(1/3)-n^(1/3)` converge?

Answer 1

As `n->oo` the individual terms tend to `0`, but the sum tends to `oo`.

Explanation:

Given:

`a_n = (n+1)^(1/3)-n^(1/3)`

Use the difference of cubes identity:

`a^3-b^3 = (a-b)(a^2+ab+b^2)`

with `a=(n+1)^(1/3)` and `b=n^(1/3)` as follows:

`lim_(n->oo) a_n = lim_(n->oo) ((n+1)^(1/3)-n^(1/3))`

`color(white)(lim_(n->oo) a_n) = lim_(n->oo) (((n+1)^(1/3)-n^(1/3))((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3)))/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))`

`color(white)(lim_(n->oo) a_n) = lim_(n->oo) ((n+1)-n)/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))`

`color(white)(lim_(n->oo) a_n) = lim_(n->oo) 1/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))`

`color(white)(lim_(n->oo) a_n) = 0`

We also find:

`sum_(n=1)^N a_n = sum_(n=1)^N ((n+1)^(1/3)-n^(1/3))`

`color(white)(sum_(n=1)^N a_n) = sum_(n=1)^N (n+1)^(1/3)-sum_(n=1)^N n^(1/3)`

`color(white)(sum_(n=1)^N a_n) = sum_(n=2)^(N+1) n^(1/3)-sum_(n=1)^N n^(1/3)`

`color(white)(sum_(n=1)^N a_n) = (N+1)^(1/3)+color(red)(cancel(color(black)(sum_(n=2)^N n^(1/3))))-color(red)(cancel(color(black)(sum_(n=2)^N n^(1/3)))) - 1^(1/3)`

`color(white)(sum_(n=1)^N a_n) = (N+1)^(1/3)-1`

Hence:

`sum_(n=1)^oo a_n = lim_(N->oo) ((N+1)^(1/3)-1) = oo`