Thread: Harmonic Numbers

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

Prove that for , the sum is not an integer.

Moderator approved CB

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.

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 is indeed not an integer ( , and also .

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 not integer" to conclude this: it must be obvious that it is not an integer since for ...and

thus your conclusion doesn't follow: for example, aren't integers, but 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 . Choose now , and now divide and sum and stuff.

Tonio

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.