Question

How do you find a formula for the sum n terms `Sigma1/n^3(i-1)^2` and then find the limit as `n->oo`?

Answer 1

`sum_(i=1)^n 1/n^3(i-1)^2 = 1/(6n^2)(2n-1)(n-1)`

`lim_(n rarr oo)sum_(i=1)^n 1/n^3(i-1)^2 = 1/3`

Explanation:

Let `S_n = sum_(i=1)^n 1/n^3(i-1)^2`
`:. S_n = 1/n^3 sum_(i=1)^n (i^2-2i+1)`
`:. S_n = 1/n^3 { sum_(i=1)^n i^2 - 2sum_(i=1)^ni + sum_(i=1)^n1 }`

And using the standard results:
`sum_(r=1)^n r = 1/2n(n+1)` and `sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)`

We have;

`S_n = 1/n^3 { 1/6n(n+1)(2n+1) -(2)1/2n(n+1) + n }`
`:. S_n = n/(6n^3) { (n+1)(2n+1) - 6(n+1) + 6 }`
`:. S_n = 1/(6n^2) { 2n^2+3n+1-6n-6+6 }`
`:. S_n = 1/(6n^2) { 2n^2-3n+1 }`
`:. S_n = 1/(6n^2)(2n-1)(n-1)`

Now we examine the behaviour of `S_n` as `n rarr oo`.
We have;

`S_n = 1/(6n^2) { 2n^2-3n+1 }`
`:. S_n = 1/3 -1/(2n)+1/(6n^2)`
`:. lim_(n rarr oo)S_n = lim_(n rarr oo) { 1/3 -1/(2n)+1/(6n^2) }`
`:. lim_(n rarr oo)S_n = lim_(n rarr oo) (1/3) - lim_(n rarr oo)(1/(2n)) + lim_(n rarr oo)(1/(6n^2))`

And as both `1/n rarr 0` and `1/n^2 rarr 0` as `n rarr oo`, we have;

`:. lim_(n rarr oo)S_n = 1/3 - 0 + 0`
`:. lim_(n rarr oo)S_n = 1/3`