Last one:
ai=n(ai-1+1)
Hence ai-1+1/ai-1 = ai/nai-1
hence,
(a
i+1)/(a
i) = a
n/n!a
1 = a
n/n!
You can prove that an = n+nan-1 = n+n(1+an-2) = ...=n2+n = n(n+1)
Hence The required limit is
n
inf n+1/(n-1)!
Now invoke the inequality, n!>2n
Hence 0<n+1/(n-1)! < n+1/2n-1
By Sandwich Theorem, the required limit is zero
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
946
