use mathematical induction to prove: for any interger n (greater than or equal to) 2,...?

Question

1/(n+1) + 1/(n+2) + 1/(n+3) + . . . + 1/(2n) > 13/24

I've done the base case (true) and stated the induction hypothesis, but I'm stuck on how to prove this.

Answer 1

Assume that for some integer n >= 2,

S_n = 1/(n+1) + 1/(n+2) + 1/(n+3) + . . . + 1/(2n) > 13/24

Now we must show that

S_n+1 = 1/(n+2) + 1/(n+2) + 1/(n+3) + . . . + 1/(2n+2) > 13/24

We have

S_n+1 - S_n = 1/(2n+1) + 1/(2n+2) - 1/(n+1)
= 1/(2n+1) + 1/(2n+2) - 2/(2n+2)
= 1/(2n+1) - 1/(2n+2)
> 0.

Therefore, S_n+1 > S_n > 13/24.

Answer 2

Comparison theorem.

Substitute 2n for all the denominators. This is a lower limit.
The sum is n* (1/(2*n)) or 1/2 or 12/24

All values of this series are greater than 12/24 for all n greater than 1.

Are you sure about the 13/24?