Problem 48

ThePerfectHacker · #1 $Old$ 04-10-2008, 11:20 PM

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

Math's-only-a-game · #2 $Old$ 04-21-2008, 03:09 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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

Let k ÎZ 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.

icemanfan · #3 $Old$ 04-23-2008, 04:14 PM

By the alternating series theorem, the partial sum will always be less than one but greater than zero, and therefore not an integer.

Aryth · #4 $Old$ 04-25-2008, 02:31 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

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

The series you presented is the Alternating Harmonic Series, which is Conditionally Convergent, the series is represented by:

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

The series' terms look like such:

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

This series converges to $\ln{2}$

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

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

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

Danshader · #5 $Old$ 04-25-2008, 02:36 PM

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

Aryth · #6 $Old$ 04-25-2008, 03:14 PM

Yeah, that is a distinct possibility...

The Alternating Series does equal:

H(n) - H(2n)

Where H(n) is the n-th harmonic number

Henderson · #7 $Old$ 05-14-2008, 08:11 AM

Huh. I stayed away from this one because I didn't pick up on the series alternating- I read $\pm \frac{1}{n}$ as saying each term could either be added or subtracted, without nessecarily alternating.

Is there a similar solution to this problem?

Aryth · #8 $Old$ 06-04-2008, 12:01 PM

You can't know what sign the last number of the series is going to be, that all depends on n, so the $\pm$ means that it can be positive or negative depending on n. The initial pattern reveals an alternating series.

Jacobsen · #9 $Old$ 06-06-2008, 02:14 PM

[FONT='Cambria Math','serif']My first thought was to try an inductive argument, but I had a lot of difficulty getting it going. I don’t 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

n≥2; 1-1/2+1/3-…±1/n ∈ (0,1).

Let Pn denote the proposition that

1-1/2+1/3-…±1/(n-1) ∈ (0,1)

and

1-1/2+1/3-…±1/(n-1)±1/n∈ (0,1).

Then P3 is true since

1-1/2=1/2∈ (0,1)

and

1-1/2+1/3=5/6∈ (0,1)

Assume Pn is true and that n is even. Then

1-1/2+1/3-…+1/(n-1) ∈ (0,1)

and

1-1/2+1/3-…+1/(n-1)-1/n ∈ (0,1).

Because 1/(n+1) < 1/n, it follows from the inductive hypothesis that

1-1/2+1/3-…+1/(n-1)-1/n+1/(n+1) ∈ (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. //

[/font]

meymathis · #10 $Old$ 07-07-2008, 05:14 PM

Quote:

Originally Posted by Danshader $View Post$

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

These two series do not converge.