# cal problem #2

• Jun 1st 2006, 04:20 PM
Nichelle14
cal problem #2
if n is greater than or equal to 2, show that

1/2 + 1/3 + ... + 1/n is less than or equal to ln n is less than or equal to 1 + 1/2 + 1/3 + ... + 1/ n-1
• Jun 1st 2006, 08:08 PM
malaygoel
:)
Quote:

Originally Posted by Nichelle14
if n is greater than or equal to 2, show that

1/2 + 1/3 + ... + 1/n is less than or equal to ln n is less than or equal to 1 + 1/2 + 1/3 + ... + 1/ n-1

to prove:
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 +..........
hence e^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 :)