A(k)=k!*(1 + k+1 + (k+1)(k+2)) = k!*(k+2)^2
Then the nth term of B is 1/(n!*(n+2))
The nth term of B is <= 1/2^(n+1)
** (equal when n=2)
so limit B(n) <= 1/4 + 1/8 + 1/16 + . . . = 1/2 = .5
The nth term of B is also >= 1/3^n
** (equal when n=1)
so limit B(n) >= 1/3 + 1/9 + 1/27 + . . . = 1/2 = .5
and thus limit B(n)=.5

Posted by xdog
on 20150804 10:30:53 