Question
Here's where my proof starts:
I know the summation of n is 1+(2/n). Somewhere after this, I my math leads me to contradict the claim that P(n) implies P(n+1). Please show steps from here to the end of the proof so I can see where I am going wrong or not manipulating the inequality correctly.
Explanation / Answer
You have read the problem incorrectly, its not sum of (1/2^n) this series converges to 2 but the problem is about 1/n which diverges So here is the proof to prove it by induction, first we verify the base case n = 1: 1/1 + 1/2 >= 1+1/2 as required. now assume that n = k is true ie the property is true for n = k ie: 1/1+ 1/2+ ... +1/(2^k) >= 1 + k/2 so to show it for n = (k+1) ie 1/1+ 1/2+ ... +1/(2^k) + 1/((2^k)+1) + ... + 1/(2^(k+1)) >= 1+(k+1)/2 so, we have to show that 1/((2^k)+1) + ... + 1/(2^(k+1)) >= 1/2 hence by adding 1/1+ 1/2+ ... +1/(2^k) >= 1 + k/2 and 1/((2^k)+1) + ... + 1/(2^(k+1)) >= 1/2 we get the required result so, to show 1/((2^k)+1) + ... + 1/(2^(k+1)) >= 1/2 we just have to observe that the expression on left has exactly 2^k terms and that each term is greater than equal to 1/(2^(k+1)) ( As i 1/j i,j positive). so 1/((2^k)+1) + ... + 1/(2^(k+1))>=1/(2^(k+1))+1/(2^(k+1))...1/(2^(k+1))(2^k times) == 2^k/(2^(k+1)) = 1/2. hence proved. So by induction we have the required result