Question

How you solve this? `lim_(n->oo)prod_(k=1)^n((k+1)^2)/(k(k+2))`

Answer 1

`lim_(n rarr oo) prod_(k=1)^n ((k+1)^2)/(k(k+2)) = 2`

Explanation:

We can find an exact formula for the product as follows:

`prod_(k=1)^n ((k+1)^2)/(k(k+2))`
`" " = (1+1)^2/(1(1+2)) * (2+1)^2/(2(2+2)) * (3+1)^2/(3(3+2)) * ... * (n+1)^2/(n(n+2))`

`" " = (2)^2/(1(3)) * (3)^2/(2(4)) * (4)^2/(3(5)) * ... * (n+1)^2/(n(n+2))`

`" " = ( 2*3*4 * ... (n+1) )^2 / ((1*2*3 * ... * n)(3*4*5 * ... (n+2)))`

`" " = (( 1*2*3*4 * ... (n+1) )^2 (1*2))/ ((1*2*3 * ... * n)(1*2*3*4*5 * ... (n+2)))`

`" " = (2(n+1)!(n+1)!)/ ((n!)(n+2)!)`

`" " = (2*n!(n+1)(n+1)!)/ ((n!)(n+1)!(n+2))`

`" " = (2(n+1))/(n+2)`

And so:

`lim_(n rarr oo) prod_(k=1)^n ((k+1)^2)/(k(k+2)) = lim_(n rarr oo) (2(n+1))/ ((n+2))`
`" " = lim_(n rarr oo) (2n+2)/(n+2)`

`" " = lim_(n rarr oo) (2n+2)/(n+2) *(1/n)/(1/n)`

`" " = lim_(n rarr oo) (2+2/n)/(1+2/n)`

`" " = (2+0)/(1+0)`

`" " = 2`

As `lim_(n rarr oo) (2/n) =0`