to prove:Originally Posted byNichelle14

1/2 + 1/3 + ... + 1/n <= ln n <= 1 + 1/2 + 1/3 + ... + 1/n-1

let 1/2 + 1/3 + ... + 1/n= a

ln n =b

1+ 1/2 + 1/3 + ... + 1/n-1=c

all a,b,c are positive

hence the question is equivalent to e^a <=e^b <= e^c

proof of e^b<= e^c

e^b =n

we know that, e^x= 1 + x+ some positive quantity

hence,e^x > 1+x(x is a positive number)

now, e^c =e^(1 + 1/2 + 1/3 + ... + 1/n-1)

=e^1*e^(1/2)...*e^(1/(n-1))

>(1+1)(1+1/2)....(1+1/(n-1))

>2*3/2*4/3.........*n/(n-1)

>n=e^b

proof of e^a<= e^b

for x<1 e^x < 1+x+x^2 + x^3 +..........

hencee^x < 1/(1-x)

e^a= e^(1/2)e^(1/3)e^(1/4).....e^(1/n)

<1/(1-1/2)*1/(1-1/3)*1/(1-1/4)*.......*1/(1-1/n)

< 2*3/2*4/3*.....*n/(n-1)

<n=e^b