Question

How do you find the limit of `s(n)=64/n^3[(n(n+1)(2n+1))/6]` as `n->oo`?

Answer 1

`lim_(n->oo)s(n)=lim_(n->oo)64/n^3[(n(n+1)(2n+1))/6]=21 1/3`

Explanation:

`lim_(n->oo)s(n)=lim_(n->oo)64/n^3[(n(n+1)(2n+1))/6]`

= `lim_(n->oo)64/6[n/n*(n+1)/n*(2n+1)/n]`

= `lim_(n->oo)64/6[1*(1+1/n)*(2+1/n)]`

= `64/6*1*1*2`

= `64/3=21 1/3`