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 PMThePerfectHackerProblem 48
1) Let $\displaystyle n\geq 2$ prove that $\displaystyle 1 - \frac{1}{2}+\frac{1}{3} - ... \pm \frac{1}{n}$ is not an integer.

- Apr 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)

- Apr 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.

- Apr 25th 2008, 12:31 PMAryth
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 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 - Apr 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 $\displaystyle \pm \frac{1}{n}$ 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 AMAryth
You can't know what sign the last number of the series is going to be, that all depends on n, so the $\displaystyle \pm$ means that it can be positive or negative depending on n. The initial pattern reveals an alternating series.

- Jun 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] - Jul 7th 2008, 03:14 PMmeymathis
- Jul 7th 2008, 08:06 PMmeymathis
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 AMLore
- Sep 8th 2008, 09:59 AMbkarpuz
As meymathis said, these two series do not converge, in other words, $\displaystyle A=\infty$ and $\displaystyle B=\infty$.

So you do algebric operations on infinite numbers, which may confuse your mind. - Oct 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 - Oct 12th 2008, 07:48 AMshawsend
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.