1) Let $\displaystyle n\geq 2$ prove that $\displaystyle 1 - \frac{1}{2}+\frac{1}{3} - ... \pm \frac{1}{n}$ is not an integer.

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- Apr 10th 2008, 09:20 PM #1

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- Apr 21st 2008, 01:09 PM #2

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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.

- Apr 23rd 2008, 02:14 PM #3

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- Apr 25th 2008, 12:31 PM #4
The series you presented is the Alternating Harmonic Series, which is Conditionally Convergent, the series is represented by:

$\displaystyle \sum_{n=1}^{\infty} \left(\frac{(-1)^{n+1}}{n}\right)$

The series' terms look like such:

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - ... \pm \frac{1}{n}$

This series converges to $\displaystyle \ln{2}$

Since the series converges to $\displaystyle \ln{2}$ and since:

$\displaystyle |a_{n+1}| < |a_n|$

Then for $\displaystyle n \geq 2$ the series can never reach one since it is incrementing up or down by smaller amounts. Since you subtract $\displaystyle \frac{1}{2}$ 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 $\displaystyle n \geq 2$ because all terms are decreasing,therefore the partial sums remain between 1 and 0.

- Apr 25th 2008, 12:36 PM #5

- Apr 25th 2008, 01:14 PM #6

- May 14th 2008, 06:11 AM #7
Huh. I stayed away from this one because I didn't pick up on the series alternating- I read $\displaystyle \pm \frac{1}{n}$ as saying each term could either be added or subtracted, without nessecarily alternating.

Is there a similar solution to this problem?

- Jun 4th 2008, 10:01 AM #8

- Jun 6th 2008, 12:14 PM #9

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[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]

- Jul 7th 2008, 03:14 PM #10

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- Jul 7th 2008, 08:06 PM #11

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Ok, how about this. It is basically Aryth's post but a bit more explicit.

Let $\displaystyle A_N = 1 - \frac{1}{2} + \ldots \pm\frac{1}{N}$ which is just the partial sums.

Consider the (sub) sequence of partial sums:

$\displaystyle

O_N = 1 - \frac{1}{2} + \ldots + \frac{1}{2N+1}

$ for $\displaystyle N\geq1$

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

$\displaystyle O_N$ is monotonically decreasing:

$\displaystyle O_{N+1}-O_N = - \frac{1}{2N+2} + \frac{1}{2N+3} < 0$

Note that $\displaystyle O_0=1$ and so $\displaystyle 1>O_N\geq\ln(2)\approx0.693$ for $\displaystyle N>0$ and so cannot be an integer.

Likewise for the partial sums:

$\displaystyle

E_N = 1 + \ldots - \frac{1}{2N}

$ for $\displaystyle N\geq1$

except that $\displaystyle E_N$ monotonically increases from 1/2 to ln(2).

Put it together and we just showed the odd and even elements of the partial sums $\displaystyle A_N$ are never integers after 1.

- Jul 30th 2008, 05:58 AM #12

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- Sep 8th 2008, 09:59 AM #13

- Oct 10th 2008, 10:49 PM #14
Suppose that . Choose an integer such that .

Then

Consider the lowest common multiple of . This number will be of the form , where is an odd integer. Now multiply both sides of the equation by this number, to get

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

is not an integer, since is odd. So the left hand side is not an integer, and hence neither is the right hand side. That means that is not an integer.

Not:

http://plus.maths.org/issue12/features/harmonic/index.html

- Oct 12th 2008, 07:48 AM #15

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I'm confussed since $\displaystyle \sum_{k=1}^{n}\frac{(-1)^{n+1}}{k}<1$

However $\displaystyle H_n=\sum_{k=1}^{n}\frac{1}{k}\to\infty$ and my understanding is that $\displaystyle H_n$ is never an integer for $\displaystyle n>1$. This one would seem to be more interesting to prove.

[edit] I think that's what Susan did. Never mind but perhaps we should make it explicit that's what's going on.