Prove that ∀n ∈ N{1},
(1/1!)+(1/2!) + (1/3!) +···+ (1/n!) < 3− [2/(n+1)!]
after suppose it is true for n = k,
so (1/1!)+(1/2!) + (1/3!) +···+ [1/ (n)!]< 3− [2/(k+1)!]
then for n= k+1, (1/1!)+(1/2!) + (1/3!) +···+ [1/ (k)!] + [1/(k+1)!]< 3− [2/(k+1)!]+[1/(k+1)!] = 3 - [1/(k+1)!]
I proved it is true for n = k+1 by 3-[1/(k+1)!] < 3-[2/(k+2)!]
Is there another way to prove it by induction? something like transform the inequality?
You can also transform the Left Hand side like this: (1/1!)+(1/2!) + (1/3!) +···+ [1/ (k)!] + [1/(k+1)!]
= 3 - [2/(k+1)!] + [1/(k+1)!]
= 3 - [1/(k+1)!]
= 3 - [(k+2)/(k+2)!]
= 3 - [2/(k+2)!] - [k/(k+1)!] < 3 - [2/(k+2)!] (what we need to prove for n= k+1)