Lonely

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15 years, 74 days

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These are answers submitted by Lonely

a[k] := (2/5)*a[k-1]*ln((1/10)*(exp(1))^10*x^10/a[k-1]^10)/ln(10^a[k-1]*(exp(1))^4)

 

@hirnyk 

 

solve((1+x)^(p+1) > 2^p*(x^p+x), p)

 

assuming 1 < x < 2

 

how to show

f(x) = m * log(x) / 2^m + (1-x^m) / (1+x)^m

 

x >= 1

 

it looks like the (i am not sure):

 

df/dx > 0 for 0 < m <=3

df/dx < 0 for m < 0

df/dx < 0 for m > 0

 

 

there was a misprint

how to see it :

 

p*(int(f(x), x = 0 .. infinity)) = sum(int(f(x), x = n .. n+p), n = 0 .. infinity)+sum((p-n)*(int(f(x), x = n-1 .. n)), n = 1 .. p-1)

 

e < (1+1/n)^((1/8)*(n^(1/3)+(n+1)^(1/3))^3)

 

thans for your reply.

Thanks a lot for your help.

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