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# how to show that..... watch

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1. anyone can help me to show that (1+1/n)^n < n , n >=3
using binomial expansion?

thanx
2. Assume that n >= 3, since otherwise the inequality isn't true.

(1 + 1/n)^n
= nCk (1/n)^k
= 2 + nCk (1/n)^k
= 2 + (1/k!)[n/n][(n - 1)/n]...[(n - k + 1)/n]
= 2 + (1/k!)[(n - 1)/n]...[(n - k + 1)/n]
<= 2 + 1/k!
= e
< 3
<= n
3. I dont think Maclaurin Series is supposed to be used in this question. In fact, it is proved in the following chapter of the book I am currently working on.

I am thinking: 1+1+1/k! < 1+1+1/2(n-1)=(1/2)n + 3/2

when n>3, (1/2)n + 3/2 < n

could anyone check whether it is a correct prove for me please?
4. 1/k!
= 1/2 + (1/6)*(1 + 1/4 + 1/(4.5) + 1/(4.5.6) + ... )
<= 1/2 + (1/6)*(1 + (1/4) + (1/4)^2 + (1/4)^3 + ... )
= 1/2 + (1/6)/(3/4)
= 13/18

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Updated: November 18, 2005
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