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# Tricky sequence question watch

1. Let {en } be the seqence by setting en = 1+1+(1/2!)+(1/3!)+(1/4!)+...+(1/n!) = Sum of k=0 to n (1/k!) for n = 0,1,2,3..., where we set 0! = 1

i) Show that {en } is a strictly increasing sequence.

ii) Explain why the following inequalities hold: en less than or equal to 1+1+(1/2)+(1/22)+(1/23 )+...+(1/2n-1 )=3 - (1/2n-1 ) less than 3.

iii) Deduce that {en } is convergent.

This question shows that it is possible to define the number e = 1+1+(1/2!)+(1/3!)+(1/4!)+...+(1/n!)+...= sum of k=0 to infinity (1/k!)

I'm really stuck on this help would be really apprieciated!!
2. For i)

en+1 = en + 1/[(n+1)!]

n>0 => 1/[(n+1)!]>0
=> en+1 > en

Therefore {en } is strictly increasing
3. (ii)
e_n = 1+1+(1/2!)+(1/3!)+(1/4!)+...+(1/n!)
= 1 + 1 + 1/2 + 1/2*3 + 1/2*3*4 + ... + 1/2*3*...*n
<= 1 + 1 + 1/2 + 1/2*2 + 1/2*2*2 + ... + 1/2*2*...*2
= 1 + 1 + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^(n-1)
= 3 - 1/(2^(n-1)) -- sum of a geometric progression
<= 3

(iii) {e_n} is increasing and bounded above, hence it must converge.

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Updated: November 18, 2005
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