Question

Lim n approaches infinity`6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=`?

lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?

Answer 1

`11`

Explanation:

After expanding

`f(n)= 6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=11+3/n-4/n^2`

so

`lim_(n->oo)f(n)=lim_(n->oo)(11+3/n-4/n^2)=11`

Answer 2

I'm guessing this is typed incorrectly.

Explanation:

The term 5n(n+1)/2n means `5n(n+1)/2n` whose limit is `oo`.

My guess is the question wants (5n(n+1))/(2n) which is `(5n(n+1))/(2n)`

Similarly, 4n(n+1)(2n+1)/6n^2 is `4n(n+1)(2n+1)/6n^2` which I suppose should be `(4n(n+1)(2n+1))/(6n^2)`

So here's what I guess the expression should be:

`6/n((2n)/3+(5n(n+1))/(2n)-(4n(n+1)(2n+1))/(6n^2))`

`= (4n)/n + (15n(n+1))/n^2 - (4n(n+1)(2n+1))/n^3`

`= 4 +15((n/n)((n+1)/n)) - 8(n/n)((n+1)/n)((2n+1)/n)`

The limit as `n rarroo` is

`4 +15(1)(1)-8(1)(1)(2) = 4+15-8 = 11`

Answer 3

`lim_(n rarr oo) 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] = 11`

Explanation:

Let `S = 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ]`
`:. S = 6[ 1/(6n^2){ 4n^2 + 15n(n+1) - 4(n+1)(2n+1) } ]`
`:. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 4(2n^2+3n+1) ]`
`:. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 8n^2-12n-4 ]`
`:. S = ( 11n^2 + 3n - 4 ) / (n^2)`
`:. S = 11 + 3/n - 4/n^2`

So:

`lim_(n rarr oo) S = lim_(n rarr oo) (11 + 3/n - 4/n^2)`
`lim_(n rarr oo) S = lim_(n rarr oo) 11 + 3lim_(n rarr oo) 1/n - 4lim_(n rarr oo)1/n^2`

`lim_(n rarr oo) S = 11 + 0 + 0`
`lim_(n rarr oo) S = 11`

This ca be verified by looking at the graph of `y=11 + 3/x - 4/x^2`
graph{11 + 3/x - 4/x^2 [-27.95, 76.05, -19, 33]}