prime infinite series

**N.A.G** · October 23rd 2008, 09:04 AM

All i really want to know is does the series 1+(1/2)+(1/3)+(1/5)+ ... +(1/n). When n is a prime , converge. and if it does can anyone explain why.

Thanks for your help
nag

**ThePerfectHacker** · October 23rd 2008, 09:43 AM

Originally Posted by N.A.G

All i really want to know is does the series 1+(1/2)+(1/3)+(1/5)+ ... +(1/n). When n is a prime , converge. and if it does can anyone explain why.

Thanks for your help
nag

The series diverges. As Euler showed.

Define $\pi (n)$ to be the primes $\leq n$ . For a given $n$ (sufficiently large) let $p_1,p_2,...,p_{\pi (n)}$ be the primes $\leq n$ .
Define $\Pi (n) = \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1}$ .

Notice that $\left( 1 - \frac{1}{p_k} \right)^{-1} = \sum_{a_k =0}^{\infty} \frac{1}{p^{a_k}}$ .

Therefore, we see that $\Pi (n) = \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1}$ .

By the fundamental theorem of arithmetic any number $1\leq m\leq n$ be be written as $m=p_1^{b_1}...p_{\pi (n)}^{b_{\pi(n)}}$ where $b_k\geq 0$ .
It follows that,
$\sum_{k=1}^n \frac{1}{k} \leq \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1} = \Pi (n)$ .

Since the harmonic series diverges it means $\lim ~ \Pi (n) = \infty \implies \lim ~ \log \Pi (n) = \infty$ .

However,
$\log \Pi (n) = \log \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1} = - \sum_{k=1}^{\pi(n)} \log \left( 1 - \frac{1}{p_k} \right) = \sum_{k=1}^{\pi (n)} \sum_{j=1}^{\infty}\frac{1}{jp_k^j}$

Thus,
$\log \Pi (n) = \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right) + \sum_{k=1}^{\pi (n)} \sum_{j=2}^{\infty}\frac{1}{jp_k^j}$

Notice that,
$\sum_{j=2}^{\infty} \frac{1}{jp_k^j} \leq \sum_{j=2}^{\infty} \frac{1}{p_k^j} = \frac{1}{p_k^2(1-\tfrac{1}{p_k})} \leq \frac{2}{p_k^2}$

We have shown,
$\log \Pi (n) \leq \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right) + 2 \left( \frac{1}{p_1^2}+...+\frac{1}{p_{\pi(n)}^2} \right) \leq \sum_{k=1}^{\pi(n)} \frac{1}{p_k} + 2 \sum_{j=1}^{n} \frac{1}{j^2}$

Remember that $\sum_{j=1}^{\infty} \frac{1}{j^2} < \infty$ .

What this means is that if $\sum_{k=1}^{\infty} \frac{1}{p_k} < \infty$ then $\log \Pi (n)$ would be bounded.
This is impossible, it is not bounded.

Thus, the sum of prime reciprocals must diverge.

**NonCommAlg** · October 23rd 2008, 05:58 PM

a very clever proof of this result is due to Paul Erdos. see "Second Proof" in here. note that in the "Lower estimate" part of the proof the inclusion $\subset$ is actually an equality.

although this doesn't change anything in the proof, but still the wikipedia author should have been more careful!

**CaptainBlack** · October 24th 2008, 11:53 PM

Originally Posted by NonCommAlg

a very clever proof of this result is due to Paul Erdos. see "Second Proof" in here. note that in the "Lower estimate" part of the proof the inclusion $\subset$ is actually an equality.

although this doesn't change anything in the proof, but still the wikipedia author should have been more careful!

It's Wikipedia, if you don't like the article or its wrong it's your job to change it!

CB

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