Question

How do you use the summation formulas to rewrite the expression `Sigma (6k(k-1))/n^3` as k=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Answer 1

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

Explanation:

Let `S_n = sum_(k=1)^n 1/n^3 6k(k-1)`
`:. S_n = 6/n^3sum_(k=1)^n (k^2-k)`
`:. S_n = 6/n^3{sum_(k=1)^n (k^2) - sum_(j=1)^n (k)}`

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

We have;

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

And this has been calculated using Excel for `n=10, 100, 1000, 10000`

What happens as `n rarr oo`?

[ NB As an additional task we could possibly conclude that as `n rarr oo` then `S_n rarr 2`; This is probably the conclusion of this question]

Now, `S_n = 2/n^2 (n^2-1)`

`:. S_n = 2n^2/n^2 - 2/n^2`
`:. S_n = 2 - 2/n^2`

And so,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (2 - 2/n^2)`
`:. lim_(n rarr oo) S_n = lim_(n rarr oo) (2) - lim_(n rarr oo)(2/n^2)`
`:. lim_(n rarr oo) S_n = 2 + 2 lim_(n rarr oo)(1/n^2)`
`:. lim_(n rarr oo) S_n = 2`, `" as "(lim_(n rarr oo)(1/n^2) =0)`
Which confirms our assumption!