# number theory proof

• Sep 13th 2008, 09:53 AM
rmpatel5
number theory proof
• Sep 13th 2008, 10:00 AM
rmpatel5
number theory proof
• Sep 13th 2008, 11:52 AM
PaulRS
Quote:

Originally Posted by rmpatel5

Let $
p_n
$
be the greatest prime not exceeding $n$.

So we must have: $
p_n \leqslant n < 2 \cdot p_n
$
(1) . Suppose this were false, then $
p_n \leqslant 2 \cdot p_n \leqslant n
$
but, by Bertrand's Postulate there's a prime $p$ such that $
p_n < p < 2 \cdot p_n
$
which is absurd since $
p_n
$
is the greatest prime not exceeding $n$.

Now suppose $
{1 + \tfrac{1}
{2} + ... + \tfrac{1}
{{p_n }} + ... + \tfrac{1}
{n}} \in
\mathbb{Z}^ +

$
then $a=
\tfrac{{n!}}
{{p_n }} \cdot \left( {1 + \tfrac{1}
{2} + ... + \tfrac{1}
{{p_n }} + ... + \tfrac{1}
{n}} \right) \in \mathbb{Z}^ +
$
since $
\tfrac{{n!}}
{{p_n }}
$
is an integer

$
a = \underbrace {\tfrac{{n!}}
{{p_n }} \cdot \tfrac{1}
{1} + \tfrac{{n!}}
{{p_n }} \cdot \tfrac{1}
{2} + ...}_{ \in \mathbb{Z}} + \tfrac{{n!}}
{{p_n }} \cdot \tfrac{1}
{{p_n }} + \underbrace {... + \tfrac{{n!}}
{{p_n }} \cdot \tfrac{1}
{n}}_{ \in \mathbb{Z}}
$

By (1) we have that $
p_n ^2
$
doesn't divide $n!$ so $a$ is not an integer. Absurd
• Sep 13th 2008, 01:10 PM
Aryth
There is a classical proof provided by Nicole Oresme, where you consider the subsequence $\{H_{2^k}\}^{\infty}_{k=0}$:

$H_1 = 1 = 1 + 0\left(\frac{1}{2}\right)$

$H_2 = 1 + \frac{1}{2} = 1 + 1\left(\frac{1}{2}\right)$

$H_4 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right)$

$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) = 1 + 2\left(\frac{1}{2}\right)$

$H_8 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right)$

$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) = 1 + 3\left(\frac{1}{2}\right)$

And therefore, in general:

$H_{2^k} \geq 1 + k\left(\frac{1}{2}\right)$.

And since the subsequence $\{H_{2^k}\}$ is unbounded, the sequence $\{H_n\}$ diverges.

And therefore it is sufficient to say that if the sequence diverges, no integer can be reached and therefore:

$\{H_n\} \notin \mathbb{Z}$
• Sep 13th 2008, 02:10 PM
awkward
Quote:

Originally Posted by Aryth
There is a classical proof provided by Nicole Oresme, where you consider the subsequence $\{H_{2^k}\}^{\infty}_{k=0}$:

$H_1 = 1 = 1 + 0\left(\frac{1}{2}\right)$

$H_2 = 1 + \frac{1}{2} = 1 + 1\left(\frac{1}{2}\right)$

$H_4 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right)$

$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) = 1 + 2\left(\frac{1}{2}\right)$

$H_8 = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right)$

$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) = 1 + 3\left(\frac{1}{2}\right)$

And therefore, in general:

$H_{2^k} \geq 1 + k\left(\frac{1}{2}\right)$.

And since the subsequence $\{H_{2^k}\}$ is unbounded, the sequence $\{H_n\}$ diverges.

And therefore it is sufficient to say that if the sequence diverges, no integer can be reached and therefore:

$\{H_n\} \notin \mathbb{Z}$

Ahem. I have to say I don't follow you there.

You have shown that 1 + 1/2 + 1/3 + ... + 1/n can be made arbitrarily large by taking n large enough. But how do you know it can't be an integer?
• Sep 13th 2008, 02:19 PM
o_O
$S = 1 + 2 + 3 + \hdots + n \qquad n \in \mathbb{Z^{+}} \ \ \Rightarrow \ \ S \in \mathbb{Z^{+}}$

The sequence $\{n\}$ diverges but you can't conclude S isn't an integer for all $n$.
• Sep 13th 2008, 05:17 PM
ThePerfectHacker
It is also here.