Question

How do you find a formula for the sum n terms `sum_(i=1)^n (16i)/n^3` and then find the limit as `n->oo`?

Answer 1

`sum_(i=1)^n (16i)/n^3 = 18/n^2(n+1)`

`lim_(n rarr oo)sum_(i=1)^n (16i)/n^3 = 0`

Explanation:

Let `S_n = sum_(i=1)^n (16i)/n^3`
`:. S_n = 16/n^3 sum_(i=1)^n i`

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

We have;

`S_n = 16/n^3 { 1/2n(n+1) }`
`:. S_n = 18/n^2(n+1)`

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

`S_n = 18/n^2(n+1)`
`:. S_n = 18/n+18/n^2`
`:. lim_(n rarr oo)S_n = lim_(n rarr oo) { 18/n+18/n^2 }`
`:. lim_(n rarr oo)S_n = 18lim_(n rarr oo) (1/n) + 18lim_(n rarr oo)(1/n^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 = 0 + 0`
`:. lim_(n rarr oo)S_n = 0`