3 Answers | Add Yours
You should perform mathematical induction method, hence, you should consider the two steps of the method.
You need to start with initial step to verify if the given expression holds for n=1 such that:
`P(1): 1/1 >= 1/2 ` valid
Since the initial step is checked, you may move to the next step called inductive step, hence, you need to prove the following statement such that:
for any k, `P(k)` true, then `P(k+1)` is also true
`P(k): 1/k + 1/(k+1) + ...+ 1/(2k) >= 1/2 ` true
You need to prove that `P(k+1)` is also true such that:
`P(k+1): 1/(k+1) + 1/(k+2) + 1/(2k)+ 1/(2k+1)+ 1/(2k+2) >= 1/2`
Notice that the sum `1/(k+1) + 1/(k+2) + 1/(2k)` may be substituted by `1/2 - 1/k` , using the statement P(k), that is true.
`P(k+1): 1/2 - 1/k + 1/(2k+1) +1/(2(k+1)) >= 1/2`
`P(k+1): 1/(2(k+1)) + 1/(2k+1)- 1/k >= 0`
`P(k+1): (2k^2+k + 2k^2 + 2k - 4k^2 - 6k - 2)/(2k(k+1)) >= 0`
`P(k+1): -(3k+2)/(2k(k+1)) >= 0` invalid
Notice that the last inequality is invalid, hence, the inequality does not exist for all integers n.
I would like to ask, what if n is a natural number. Is this inequality exists?
Thank you very much ;)
We’ve answered 319,175 questions. We can answer yours, too.Ask a question