1

Let $n$ and $m$ be positive integers. Show that $\sum_{i=n}^{n+m}\frac{1}{i}=\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n+m}$ is not an integer using Bertrand's postulate.

There are a few posts asking a similar question (e.g. How do I prove this sum is not an integer) but none of them use Bertrand's postulate.

Firstly, I'm not sure if you can use it. Say that $n=8$ and $m=2$, then the sum is, $$\frac{1}{8}+\frac{1}{9}+\frac{1}{10}$$ and there are no primes in the denominator as I would like to use.

Shean
  • 877

1 Answers1

3

This answer is thanks to Gerry Myerson's insight.

There are 2 possibilities, (1) $n\le m$. Bertrand's postulate applies, and use the reasoning from Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

Possibility (2), $n>m$. We have, the sum, $$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n+m}\le\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n-1}< \frac{1}{n}+\frac{1}{n}+\dots+\frac{1}{n}=\frac{n}{n}=1$$ hence it is not an integer.

Shean
  • 877