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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?

Haoran Chen
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    ??? Having proved it by induction, why would you want another way to prove it by induction? ("Proof by induction" is pretty much "cut and dried"- I can't imagine there being two different ways to prove any given statement by induction.) – user247327 Apr 13 '20 at 17:41
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    I just wondering if there is another way to prove this inequality. you can say no if you don't have another solution. please read my question not just look at the first and last sentence – Haoran Chen Apr 13 '20 at 17:43
  • Welcome to Mathematics Stack Exchange. $1/1!+1/2!+1/3!+\cdots1/n!<e-1<2\le3-2/(n+1)!$ for $n\ge1$ – J. W. Tanner Apr 13 '20 at 17:53

1 Answers1

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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)

Uyen Le
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