Problem 48 - Page 2

meymathis · #11 $Old$ 07-07-2008, 10:06 PM

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

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

Consider the (sub) sequence of partial sums:

$O_N = 1 - \frac{1}{2} + \ldots + \frac{1}{2N+1}$ for $N\geq1$

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

$O_N$ is monotonically decreasing:

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

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

Likewise for the partial sums:

$E_N = 1 + \ldots - \frac{1}{2N}$ for $N\geq1$

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

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

Lore · #12 $Old$ 07-30-2008, 07:58 AM

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

A - 2B = 0? Considering B = A/2...

bkarpuz · #13 $Old$ 09-08-2008, 11:59 AM

Quote:

Originally Posted by Lore $View Post$

A - 2B = 0? Considering B = A/2...

As meymathis said, these two series do not converge, in other words, $A=\infty$ and $B=\infty$ .
So you do algebric operations on infinite numbers, which may confuse your mind.

Suzan · #14 $Old$ 10-11-2008, 12:49 AM

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.

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

shawsend · #15 $Old$ 10-12-2008, 09:48 AM

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.

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

However $H_n=\sum_{k=1}^{n}\frac{1}{k}\to\infty$ and my understanding is that $H_n$ is never an integer for $n>1$ . 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.

Suzan · #16 $Old$ 10-12-2008, 02:51 PM

Quote:

Originally Posted by shawsend $View Post$

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

I thought that way. I searched harmonic series.

And I think this
In perfect harmony by John Webb

http://plus.maths.org/issue12/featur...nic/index.html

very good expressed verbally or in writing