Question

Evaluate the sum `sum_(i=1)^n (12i^2(i-1))/n^4` for `n=10,100,1000` and `10000`?

Answer 1

`{: (n=10, => "sum"=3.168), (n=100, => "sum"=3.019698), (n=1000, => "sum"=3.001996998), (n=10000, => "sum"=3.00019997) :}`

Explanation:

Let:

`S_n = sum_(i=1)^n (12i^2(i-1))/n^4`
`" " = 12/n^4sum_(i=1)^n {i^3-i^2}`
`" " = 12/n^4{sum_(i=1)^n i^3 - sum_(i=1)^n i^2}`

And using the standard results:

`sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)`
`sum_(r=1)^n r^3 = 1/4n^2(n+1)^2`

We have;

`S_n = 12/n^4{1/4n^2(n+1)^2 - 1/6n(n+1)(2n+1) }`
`" " = 12/n^4{1/4n^2(n+1)^2 - 1/6n(n+1)(2n+1) }`
`" " = 1/n^3{3n(n+1)^2 - 2(n+1)(2n+1) }`
`" " = (n+1)/n^3{3n(n+1) - 2(2n+1) }`
`" " = (n+1)/n^3{3n^2+3n - 4n - 2 }`
`" " = (n+1)/n^3{3n^2- n - 2 }`
`" " = {(n+1)(n-1)(3n+2) }/n^3`

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

Conclusion: What happens as `n rarr oo`?

[ NB As an additional task we could possibly conclude that as `n rarr oo` then `S_n rarr 1`; This is probably the conclusion of this question]

From the above results it looks as if:

`S_n ~~ 3` as `n rarr oo`

Let us see if this is actually the case. We can manipulate `S_n` as follows;

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

And so,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (3+ 2/n - 3/n^2 - 2/n^3)`
`" " = 3+ 0 - 0 - 0`
`" " = 3`

Which confirms our numerical observations!