How do you find the limit of #30 - (8/(6n^2)(n+1)(2n+1)) - ((4/n)(n+1))# as n approaches #oo#?

Question

1206 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-31T07:31:52

#L_(n->oo)[3-(8/(6n^2)(n+1)(2n+1))-(4/n(n+1))]=-11/3#

#L_(n->oo)[3-(8/(6n^2)(n+1)(2n+1))-(4/n(n+1))]# can be written as

#L_(n->oo)[3-(8/6(1+1/n)(2+1/n))-(4(1+1/n))]#

We know that as #n->oo#, #1/n->0#

Hence #L_(n->oo)[3-(8/6(1+1/n)(2+1/n))-(4(1+1/n))]#

= #[3-(8/6(1+0)(2+0))-(4(1+0))]#

= #(3-(8/6*1*2)-4)=3-8/3-4=-11/3#

How do you find the limit of #30 - (8/(6n^2)*(n+1)(2n+1)) - ((4/n)*(n+1))# as n approaches #oo#?