Can someone help me proove this problem by induction

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

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

thanks

This inequality becomes narrower and narrower as n grows. Therefore, to prove it by induction, you must strengthen the induction hypothesis to:

1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n - 1) - 1/(2n) = 1/(n + 1) + 1/(n + 2) + ... + 1/(2n - 1) + 1/(2n). (*)

This proves the inequality as it shows that the difference between the left and right sides of the inequality is 1/(2n) ≥ 0.

(*) is true for n = 1 as both the left and right-hand sides of (*) equal 1/2. Now, assuming that it is true for n = m, we need to prove it for n = m + 1. But then

1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n - 1) - 1/(2n)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2m - 1) - 1/(2m) + 1/(2m + 1) - 1/(2m + 2)
= [ 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2m - 1) - 1/(2m) ] + 1/(2m + 1) - 1/(2m + 2)
= [ 1/(m + 1) + 1/(m + 2) + ... + 1/(2m - 1) + 1/(2m) ] + 1/(2m + 1) - 1/(2m + 2),
by the induction hypothesis of (*) with n = m
= 1/(m + 2) + ... + 1/(2m - 1) + 1/(2m) + 1/(2m + 1) + [ 1/(m + 1) - 1/(2m + 2) ],
reordering and regrouping the terms in the sum
= 1/(m + 2) + ... + 1/(2m - 1) + 1/(2m) + 1/(2m + 1) + 1/(2m + 2).
= 1/(n + 1) + ... + 1/(2n - 1) + 1/(2n).

This proves that (*) is true for n = m + 1, so, by induction, it is true for all n.