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

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- August 6th 2010, 10:24 PMmeleseHarmonic 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 - August 6th 2010, 10:46 PMAlso 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. - August 6th 2010, 11:10 PMmelese
- August 7th 2010, 08:21 AMtonio

Uuh?? We don't need the preceeding part to "...hence not integer" to conclude this: it must bethat it is not an integer since for ...and__obvious__

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 - August 23rd 2010, 06:41 PMmadisonmay
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.