How do I prove that for every natural number n the following inequality exists.....

[1+1/3+1/5+..+1/(2n-1]/(n+1) > = [1/2+1/4+..+1/(2n)]/n

love arun

Let a_n be the sum of the first n terms of 1+1/3+1/5+...

Let b_n be the sum of the first n terms of 1/2+1/4+1/6+...

a₁ -b₁ = 1/2, and all subsequent differences are positive, so
a_n -b_n > = 1/2 for all n.

A quick look at a_n for n=1, 2, 3, 4 shows that a_n < = (n+1)/2. For higher n's, it is clear that a_n+1 -a_n < 1/2, so for all n it is true that
a_n < = (n+1)/2.

0 > = -n/2 + a_n - 1/2

(n)(a_n ) > = (n)(a_n ) -n/2 + a_n - 1/2

(n)(a_n ) > = (n+1)(a_n -1/2)

(n)(a_n ) > = (n+1)(b_n )

(1/(n+1))(a_n ) > = (1/n)(b_n )

An alternative method is to set H_n = 1 + 1/2 + ... + 1/n. Then note:

H_2n - H_n = 1/(n+1) + 1/(n+2) + ... + 1/(2n) > = 1/(2n) + 1/(4n) + ... + 1/(2n*n) = H_n /(2n)

Therefore H_2n > = H_n (1 + 1/(2n)). This can be re-arranged as:

1/(n+1) * (H_2n - H_n /2) > = 1/n(1/2 * H_n )

which gives the desired inequality on substituting.