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# exp inequality watch

1. How can I use the fact that 1+x smaller or equal to e^x,

so for every n>0, (1+1/n)^n is smaller or equal to e.....?

I know that
(1+1/n)^n=1+n(1/n)+n(n-1)(1/n)^2/2!+...+(1/n)^n=1+1+...+(1/n)^n

and when x=1, 1+1=2 is smaller or equal to e.....

But then I'm stucked.....
2. (Original post by Moon_moon)
How can I use the fact that 1+x smaller or equal to e^x,

so for every n>0, (1+1/n)^n is smaller or equal to e.....?

I know that
(1+1/n)^n=1+n(1/n)+n(n-1)(1/n)^2/2!+...+(1/n)^n=1+1+...+(1/n)^n

and when x=1, 1+1=2 is smaller or equal to e.....

But then I'm stucked.....
Go back to square one and substitute x = 1/n
3. (Original post by Kevin De Bruyne)
Go back to square one and substitute x = 1/n
thx

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