Question

How do you determine if `a_n=(1-1/8)+(1/8-1/27)+(1/27-1/64)+...+(1/n^3-1/(n+1)^3)+...` converge and find the sums when they exist?

Answer 1

The sum of the first `n` terms is given by:

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

The sum to infinity is given by:

`S = 1`

Explanation:

Let us denote the `n^(th)` term of the sequence by:

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

Then we can denote the partial finite sum by `S_n` so that:

`S_n = sum_(r=1)^n \ u_r`
`\ \ \ \ = (1-1/8)+(1/8-1/27)+... + (1/n^3-1/(n+1)^3)`

And then the infinite sum, by `S` where:

`S = sum_(r=1)^oo \ u_r`

First we can derive an expression for `S_n`, as we can write the sum of the first `n` terms as follows:

`S_n = u_1 + u_2 + ... u_n`

`\ \ \ \ \ = {1-1/8} +`
`\ \ \ \ \ \ \ \ \ \ {1/8-1/27} +`
`\ \ \ \ \ \ \ \ \ \ {1/27-1/64} +`
`\ \ \ \ \ \ \ \ \ \ vdots`
`\ \ \ \ \ \ \ \ \ \ {1/n^3-1/(n+1)^3 }`

This is a difference sum, and we can see that almost all terms will cancel with other terms:

`S_n = {1-cancelcolor(red)(1/8)} +`
`\ \ \ \ \ \ \ \ \ \ {cancelcolor(red)(1/8)-cancelcolor(blue)(1/27)} +`
`\ \ \ \ \ \ \ \ \ \ {cancelcolor(blue)(1/27)-cancel(1/64)} +`
`\ \ \ \ \ \ \ \ \ \ vdots`
`\ \ \ \ \ \ \ \ \ \ {(cancel(1/n^3)-1/(n+1)^3 }`

After which we are left with:

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

And for the sum, `S`, we have:

`S = lim_(n rarr oo) S_n`
`\ \ = lim_(n rarr oo) 1 - 1/(n+1)^3`
`\ \ = 1-0`
`\ \ = 1`