1. ## Harmonic Numbers

Challange Problem; for those who have never seen it, try this by yourselves.

Prove that for $\displaystyle n>1$, the sum $\displaystyle \displaystyle{1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}}$ is not an integer.

Moderator approved CB

2. Maybe a solution: (first thing that came to my head...)

1+1/2+...+1/n=H_n

Hence:

H_n - H_{n-1}=1/n

n(H_n - H_{n-1})=1

Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.

3. Originally Posted by Also sprach Zarathustra
Maybe a solution: (first thing that came to my head...)

1+1/2+...+1/n=H_n

Hence:

H_n - H_{n-1}=1/n

n(H_n - H_{n-1})=1

Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.
Hi Also sprach Zarathustra, you're are saying essentially that the product of an integer and a rational (that is non-integer) cannot be an integer.
To look at differently $\displaystyle H_n-H_{n-1}$ is indeed not an integer ($\displaystyle 1/n)$, and also $\displaystyle n(H_n-H_{n-1} )=n\cdot\frac{1}{n}=1$.

4. Originally Posted by Also sprach Zarathustra
Maybe a solution: (first thing that came to my head...)

1+1/2+...+1/n=H_n

Hence:

H_n - H_{n-1}=1/n

n(H_n - H_{n-1})=1

Suppose H_n integer, hence H_n - H_{n-1} not integer! BUT n(H_n - H_{n-1}) is integer (1) ==>H_n can't be integer.

Uuh?? We don't need the preceeding part to "...hence $\displaystyle H_n-H_{n-1}$ not integer" to conclude this: it must be obvious that it is not an integer since for $\displaystyle 1<n\in\mathbb{N}\,,\,\,0<\frac{1}{n}<1$ ...and

thus your conclusion doesn't follow: for example, $\displaystyle \frac{1}{3}\,,\,\frac{2}{3}$ aren't integers, but $\displaystyle 3\cdot\frac{1}{3}\,,\,3\cdot\frac{2}{3}$ are.

The proofs I know of this aren't hard but can be tricky. Hints for the most elementary one (that I know, of course):

If $\displaystyle H_n\in\mathbb{N}\,,\,\,then\,\,\,H_n\geq 2$ . Choose now $\displaystyle k\in\mathbb{N}\,\,\,s.t. \,\,\,1<2^k\leq n$ $\displaystyle \Longrightarrow l.c.m. (1,2,\ldots,n)=2^k\cdot 3^s\cdot5^r\cdot\ldots$ , and now divide and sum and stuff.

Tonio

5. Well for 1/1 + 1/2 + 1/4 +1/8 converges to two. 1/3 + 1/9 + 1/27 converges as well. This problem is really just asking if the series 1/p converges. P = prime

I am sure that this harmonic series diverges, but I really don't know how to prove it. And if it diverges, it does not converge on an integer.