Expert: Ahmed Salami - 2/17/2009

Question
Show by mathematical induction that 1/(n+1) + 1/(n+2) + ... + 1/2n > 1/2

Answer
Hi Ayodeji,
Let the series be
T(n) = 1/(n+1) + 1/(n+2) + ... + 1/2n
By mathematical induction, we have to show that for any k, T(k+1) > 1/2 whenever T(k) > 1/2
We assume that the following is true,
T(k) > 1/2
i.e
1/(k+1) + 1/(k+2) + ... + 1/2k > 1/2
We now check if, as a consequence of this, the following is true,
T(k+1) > 1/2
i.e
1/(k+1+1) + 1/(k+2+1) + ... + 1/2(k+1) > 1/2
1/(k+2) + 1/(k+3) + ... + 1/2k + 1/2k+1 + 1/2k+2 > 1/2
To confirm this, we compare T(k) and T(k+1) and find that T(k+1) is similar in terms to T(k) except that it has the terms 1/2k+1 and 1/2k+2 and not 1/k+1!!!
Now,
1/2k+1 > 1/2k+2
and so
1/2k+1 + 1/2k+2 > 1/2k+2 + 1/2k+2
1/2k+1 + 1/2k+2 > 2/2k+2
1/2k+1 + 1/2k+2 > 2/2(k+1)
1/2k+1 + 1/2k+2 > 1/(k+1)
We have thus shown that T(k+1) > T(k), and since T(k) > 1/2 it follows that T(k+1) > 1/2
That is our proof.

Normally, in mathematical induction, we start with a base case( a first number that satisfies the statement) but i skipped that because you're probably used to starting with k = 1 and this wont work in this   
case. To show this,
T(1) = 1/(1+1) = 1/2
clearly, 1/2 is not > 1/2 and so our base case in this problem would be k = 2.
The proof is therefore valid for all k > 1

You can always get back to me.
And by the way, i'm Nigerian as well.
Regards