Please see the " Microsoft word" attachement. Thank you very much.
The sum,
$\displaystyle \sum_p \frac{1}{p} $
Diverges.
I will quote Wikipedia while doing this.
This is the only proof I am familar with (I know there are many).
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You need to be familar with the Zeta function,
$\displaystyle \zeta (s)=\sum_{k=1}^{\infty} \frac{1}{n^s}$
It diverges for $\displaystyle s\leq 1$ (integral p-series test).
Converges for $\displaystyle s>1$.
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You also need to be familar with one of the most elegant formulas, "Euler-Product formula".
It states we can factorize the zeta function as,
$\displaystyle \zeta (s)=\prod_p (1-p^{-s})^{-1}$
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The final concept you need to know is how to work with infinite products. If you an infinite product of positive terms,
$\displaystyle \prod a_k$
You can take the natural logarithm to obtain,
$\displaystyle \ln \prod a_k=\ln (a_1a_2...)=\ln a_1+\ln a_2+...= \sum \ln a_k$
And then work with convergence with the infinite series!
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Now we can begin with Euler's proof.
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$\displaystyle \ln \sum_{n=1} \frac{1}{n}=\ln \prod_p (1-p^{-1})^{-1}$
By Euler's product formula.
Expressing the product as a sum we have, (natural logarithms are positive)
$\displaystyle \sum_p \ln (1-p^{-1})^{-1}$
Exponent rule for logarithms,
$\displaystyle \sum_p -\ln (1-p^{-1})$
Infinite series for logarithm ($\displaystyle |p^{-1}|<1$)
Thus,
$\displaystyle \sum_p \frac{1}{p}+\frac{1}{2p^2}+\frac{1}{3p^3}+...$
But this is (absolute convergenct thus we can rearrange),
$\displaystyle \sum_p \frac{1}{p} +\sum_p \frac{1}{p^2}\left( \frac{1}{2}+\frac{1}{3p}+... \right)$
This is strictly less than,
$\displaystyle \sum_p \frac{1}{p}+\sum_p \frac{1}{p^2} \left(1+\frac{1}{p}+\frac{1}{p^2}+... \right)$
Geomteric series ($\displaystyle |1/p|<1$),
$\displaystyle \sum_p \frac{1}{p}+\sum_p \frac{1}{p(p-1)}$
Now, if
$\displaystyle \sum_p \frac{1}{p}$
Converges.
Then, by the dominance rule,
$\displaystyle 0\leq \sum_p \frac{1}{p(p-1)}\leq \sum_p \frac{1}{p}$
Converges.
That means,
$\displaystyle \ln \sum_{n=1}^{\infty} \frac{1}{n}$
Is bounded by this sum.
Which is not possible because this is the infamous harmonic series which diverges.
Q.E.D.
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Note, since the harmonic series behaves like $\displaystyle \ln n$ and the sum of reciprical of primes is the natural logarithm of that we can say,
$\displaystyle \sum_p \frac{1}{p} \sim \ln \ln n$
(Note not a prove, an observation).
Thus, it diverges really really slowly.
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Thus the number of primes is infinite.
Because otherwise the reciprocal sum would converge.
Hi perfecthacker,
Thank you very much.
As I understand from your post, the information you gave me is belonging to the later part of this question.
Could you please show me how to do the beginning part?
a) Prove that there are exactly 2j square-free numbers whose
only prime divisors are in the set { p1, p2, ..., pj }.
b) Prove that, for sufficiently large x,
Nj(x)<= 2j * sqrt(x) (Hint: show every number can be written uniquely as the product of a perfect square and a square-free number).
(c) Prove that,
for sufficiently large x, Nj(x) < x. Why does this
mean there must be an infinite number of primes?
(d) For subsequent results, we need a little bit more. Prove that,
for sufficiently large x, Nj(x) < x2 (The proof is
exactly like the one in (c).)
Thank you very much.
This question seems difficult to me. If I misunderstand your post, please do let me know. Many thanks again.
I have forgotten to mention about this point :
The attachment in the " Microsoft word" is a proof that there are an infinite number of primes.
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Please do guide me through this question. I am completely blank for b to d. Thank you very much.
First lets do what the hint suggests (though it is pretty obvious)
Consider some positive integer x, and its prime decomposition:
$\displaystyle x=p_1 ^{a_1}p_2^{a_2}...p_m^{a_m}$,
then:
let $\displaystyle b_k = 2\,\lfloor a_k / 2 \rfloor $, and $\displaystyle c_k=a_k-b_k$, then $\displaystyle c_k=0$ or $\displaystyle c_k=1 \mbox{ for }k=1,..,m$, and $\displaystyle b_k$ is even.
Then:
$\displaystyle x=B\,C$
where $\displaystyle B=p_1 ^{b_1}p_2^{b_2}...p_m^{b_m}$ is a square, and $\displaystyle C=p_1 ^{c_1}p_2^{c_2}...p_m^{c_m}$ is square free.
Which proves the result in the suggestion.
Now the number of squares less than or equal $\displaystyle x\le \sqrt{x}$, and by part (a) the number number of square free numbers with prime factors from amoung $\displaystyle \{p_1,p_2, .. , p_j\} =2^j$ so:
$\displaystyle N_j(x)\le 2^j \sqrt(x)$
RonL
Notes:
$\displaystyle \lfloor x \rfloor $ denotes the floor function for $\displaystyle x$, that is the greatest integer less than $\displaystyle x$.
Hi Captainblack,
Thank you very much for your rely. Your answers very logical. I need your help for go on the next three part. Please! Please!
Anyhow to show that the sum of the reciprocals of the primes diverges, it's
enough to show that, for any j:
1/pj+1 + 1/pj+2 + 1/pj+3 + .... > 1/2
(e) Why is it enough to show that the above sum is greater
than 1/2 ?
(f) Show that for any x, x/pj+1 + x/pj+2 + x/pj+3 + .... >= x - Nj(x).
(Hint: Explain first why x/p is greater than or equal to the number
of numbers less than or equal to x that are divisible by p. What about
x/p + x/q where p and q are relatively prime?)
(g) Use the results from (d) and (f) to complete the proof
that the sums of the reciprocals of the primes diverge.
(e) If for any $\displaystyle j$:
$\displaystyle S_j=1/p_{j+1} + 1/p_{j+2} + 1/p_{j+3} + .... > 1/2$
then we can choose an $\displaystyle m_1$ such that:
$\displaystyle S_{j,m_1} = 1/p_{j+1} +.. 1/p_{m_1} > 1/4$,
and then:
$\displaystyle S_{m_1}=1/p_{m_1+1} + 1/p_{m_1+2} + 1/p_{m_1+3} + .... > 1/2$
then we can choose an $\displaystyle m_2$ such that:
$\displaystyle S_{m_1,m_2} = 1/p_{m_1+1} +.. 1/p_{m_2} > 1/4$,
and so on, so repeating this process $\displaystyle k$ times we have:
$\displaystyle S_j=k/4+1/p_{n+1} + 1/p_{n+2} + 1/p_{n+3} + ... = k/4 + S_n >(k+2)/4$
for any $\displaystyle k$, so we find that $\displaystyle S_j$ is greater than any number we care to name and so is divergent
RonL
I'm afraid this is not as clear as I would like it but here goes:
First the hint:
The numbers less than or equal to x divisible by p are:
$\displaystyle p, 2p, ... \lfloor x/p \rfloor p $.
There are $\displaystyle \lfloor x/p \rfloor$ of these, which is $\displaystyle \le x/p$, which is the result required by first part of the hint.
Now the number of numbers $\displaystyle \le x$ divisible by $\displaystyle p$ and or $\displaystyle q$ is less than the sum of the number $\displaystyle \le x$ divisible by $\displaystyle p$ and the number divisible by $\displaystyle q$.
So now:
$\displaystyle x/p_{j+1} + x/p_{j+2} + ....$
is greater than or equal to the number less than $\displaystyle x$ which have prime divisors are among $\displaystyle {p_{j+1}, p_{j+2}, ...}$. This final number is equal to $\displaystyle x-N_j(x)$ (as there are $\displaystyle x$ numbers less than or equal to $\displaystyle x$ and $\displaystyle N_j(x)$ of these have primes divisors among $\displaystyle p_1, .. p_j$ and no others, so:
$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x)$,
as required.
RonL
from (f) we have:
$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x)$,
from (d) we have:
$\displaystyle N_j(x)<x/2$
so:
$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x) > x-x/2=x/2$
Divide this through by $\displaystyle x$ gives:
$\displaystyle 1/p_{j+1} + 1/p_{j+2} + ...>1/2$
and (e) tells us that this means that:
$\displaystyle 1/2 + 1/3 + .. + 1/p_n + ...$
diverges.
RonL