Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Is there an elementary proof that $∑_{k=1}^n 1/k$ is never an integer?


Prove that $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is not an integer.

I tried to prove by induction on $n$, but I was stuck :(

Assume $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} = \frac{a}{b}$ for some integers $a, b$ and $a \neq b \text{and} b \neq 0$
Then $ 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n + 1} = \frac{a}{b} + \frac{1}{n + 1}$

Then how can I prove that this expression is not integer? A hint would be greatly appreciated.


share|cite|improve this question

marked as duplicate by Sivaram Ambikasaran, Jonas Meyer, Akhil Mathew Feb 28 '11 at 0:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Look at the power of 2 divisible by numerator and denominator. – Soarer Feb 28 '11 at 0:24
Your induction hypothesis is not strong enough, because simply assuming that $k$ is not an integer does not guarantee that $k+\frac{1}{n+1}$ is not an integer. So if you want to proceed by induction, you need to prove more than simply that $H_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$ is not an integer, you need to prove something about its expression as a rational written in lowest terms. – Arturo Magidin Feb 28 '11 at 0:27
This is a duplicate. Why is it open? – Eric Naslund Feb 28 '11 at 0:35
I didn't see that it was a duplicate. I have now marked it for close. – user17762 Feb 28 '11 at 0:37
up vote 3 down vote accepted

Hint: look at the largest power of 2 less than $n$. Can it get canceled out from the denominator?

share|cite|improve this answer

HINT: There is always a prime between $\frac{n}{2}$ and $n$, $\forall n \geq 4$

share|cite|improve this answer
That's overkill. – lhf Feb 28 '11 at 0:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.