Very simple question about limit superior (or upper limit)

[jwqwerty]

Homework Statement

If Sn =1/0! + 1/1! + 1/2! +... 1/n! , Tn=(1+(1/n))^n, then lim sup Tn ≤ e? (e=2.71...)

Homework Equations

1. e= Ʃ(1/n!)
2. If Sn≤Tn for n≥N, then lim sup Sn ≤ lim sup Tn

The Attempt at a Solution

By binomial theorem,
Tn= 1 + 1 + 1/(2!)(1-1/n) + 1/(3!)(1-1/n)(1-2/n) + ... + 1/n!
Hence Tn ≤ Sn＜ e,
lim sup Tn ≤ lim sup Sn＜ e

∴ lim sup Tn＜ e

But I do not get lim sup Tn ≤ e

What did i do wrong?

 

[jgens]

1. If you show that limsup T_n < e, then it follows that limsup T_n ≤ e. It turns out that it is not true that limsup T_n ＜ e (so you did mess up in coming to that conclusion), but this conclusion does not contradict the claim that T_n ≤ e.
2. If T_n ≤ S_n ＜ e, then we can conclude that limsup T_n ≤ limsup S_n ≤ e. We cannot, however, conclude that limsup S_n ＜ e based on this information; in fact, limsup S_n = e.