1. ## number theory proof

Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n

2. ## number theory proof

Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n

3. Originally Posted by rmpatel5
Let with n>1. Prove that 1 + 1/2 + 1/3 + .....1/n
Let $\displaystyle p_n$ be the greatest prime not exceeding $\displaystyle n$.

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

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

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

By (1) we have that $\displaystyle p_n ^2$ doesn't divide $\displaystyle n!$ so $\displaystyle a$ is not an integer. Absurd

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

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

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

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

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

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

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

And therefore, in general:

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

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

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

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

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

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

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

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

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

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

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

And therefore, in general:

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

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

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

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

6. $\displaystyle S = 1 + 2 + 3 + \hdots + n \qquad n \in \mathbb{Z^{+}} \ \ \Rightarrow \ \ S \in \mathbb{Z^{+}}$

The sequence $\displaystyle \{n\}$ diverges but you can't conclude S isn't an integer for all $\displaystyle n$.

7. It is also here.