# Show that for n, which is a whole number, this inequality exists. 1/(n)+1/(n+1)+...+1/(2n)>=1/2This task is from a chapter of mathematical induction, so probably it should be done by induction.

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### 3 Answers

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 ;)