1) Let prove that is not an integer.

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- April 10th 2008, 09:20 PMThePerfectHackerProblem 48
1) Let prove that is not an integer.

- April 21st 2008, 01:09 PMMath's-only-a-game
Let k ΞZ such that 2^k £ n < 2^k+1

Let m be the least common multiple of 1,2,3,…,n except 2^k.

Then multiplying S = 1 – 1/2 + 1/3 -…..± 1/n by m we have:

mS = m – m/2 + m/3 -……± m/n

Each number on the right hand side is an integer except m/2^k and hence Sm is not an integer, which implies Sm is not an integer. (Hi)

- April 23rd 2008, 02:14 PMicemanfan
By the alternating series theorem, the partial sum will always be less than one but greater than zero, and therefore not an integer.

- April 25th 2008, 12:31 PMAryth
The series you presented is the Alternating Harmonic Series, which is Conditionally Convergent, the series is represented by:

The series' terms look like such:

This series converges to

Since the series converges to and since:

Then for the series can never reach one since it is incrementing up or down by smaller amounts. Since you subtract from 1 for n=2, and since the terms are decreasing and alternating in sign, then the series will never reach one again, therefore, this can't be an integer for because all terms are decreasing,therefore the partial sums remain between 1 and 0. - April 25th 2008, 12:36 PMDanshader
how about considering:

1 +1/2 + 1/3 + 1/4 + .... ===>A

and

1/2 + 1/4 + 1/6 +.... =====>B

to get the required series:

A - 2B - April 25th 2008, 01:14 PMAryth
Yeah, that is a distinct possibility...

The Alternating Series does equal:

H(n) - H(2n)

Where H(n) is the n-th harmonic number - May 14th 2008, 06:11 AMHenderson
Huh. I stayed away from this one because I didn't pick up on the series alternating- I read as saying each term could either be added or subtracted, without nessecarily alternating.

Is there a similar solution to this problem? - June 4th 2008, 10:01 AMAryth
You can't know what sign the last number of the series is going to be, that all depends on n, so the means that it can be positive or negative depending on n. The initial pattern reveals an alternating series.

- June 6th 2008, 12:14 PMJacobsen
[FONT='Cambria Math','serif']My first thought was to try an inductive argument, but I had a lot of difficulty getting it going. I dont think what I came up with is sound, but nevertheless I decided to post what I came up with.

Proof. It suffices to show that for all

Let Pn denote the proposition thatn≥2; 1-1/2+1/3- ±1/n ∈ (0,1).

and1-1/2+1/3- ±1/(n-1) ∈ (0,1)

Then P3 is true since1-1/2+1/3- ±1/(n-1)±1/n∈ (0,1).

and1-1/2=1/2∈ (0,1)

Assume Pn is true and that n is even. Then1-1/2+1/3=5/6∈ (0,1)

and1-1/2+1/3- +1/(n-1) ∈ (0,1)

Because 1/(n+1) < 1/n, it follows from the inductive hypothesis that1-1/2+1/3- +1/(n-1)-1/n ∈ (0,1).

The case where n is odd is similar. So by the principle of mathematical induction, for all n ≥ 3, Pn is true and hence for all n ≥ 2, 1-1/2 +1/3 - ±1/n ∈ (0,1) and hence not an integer. //1-1/2+1/3- +1/(n-1)-1/n+1/(n+1) ∈ (0,1).

[/FONT] - July 7th 2008, 03:14 PMmeymathis
- July 7th 2008, 08:06 PMmeymathis
Ok, how about this. It is basically Aryth's post but a bit more explicit.

Let which is just the partial sums.

Consider the (sub) sequence of partial sums:

for

is a subsequence of which as noted above converges to ln(2) (derive using MacLauren expansion of ln at x=1). Then .

is monotonically decreasing:

Note that and so for and so cannot be an integer.

Likewise for the partial sums:

for

except that monotonically increases from 1/2 to ln(2).

Put it together and we just showed the odd and even elements of the partial sums are never integers after 1. - July 30th 2008, 05:58 AMLore
- September 8th 2008, 09:59 AMbkarpuz
As meymathis said, these two series do not converge, in other words, and .

So you do algebric operations on infinite numbers, which may confuse your mind. - October 10th 2008, 10:49 PMSuzan
Suppose that http://plus.maths.org/MI/plus/issue1...s/img-0001.png. Choose an integer http://plus.maths.org/MI/plus/issue1...s/img-0002.png such that http://plus.maths.org/MI/plus/issue1...s/img-0003.png.

Then http://plus.maths.org/MI/plus/issue1...s/img-0004.png

Consider the lowest common multiple of http://plus.maths.org/MI/plus/issue1...s/img-0005.png. This number will be of the form http://plus.maths.org/MI/plus/issue1...s/img-0006.png, where http://plus.maths.org/MI/plus/issue1...s/img-0007.png is an odd integer. Now multiply both sides of the equation by this number, to get

http://plus.maths.org/MI/plus/issue1...s/img-0008.png

Now, when multiplied out, all the terms on the left will be integers, except one:

http://plus.maths.org/MI/plus/issue1...s/img-0009.png

is not an integer, since http://plus.maths.org/MI/plus/issue1...s/img-0007.png is odd. So the left hand side is not an integer, and hence neither is the right hand side. That means that http://plus.maths.org/MI/plus/issue1...s/img-0010.png is not an integer.

Not:

http://plus.maths.org/issue12/features/harmonic/index.html - October 12th 2008, 07:48 AMshawsend