LinkBack URL
About LinkBacksAll 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.Originally Posted by N.A.G
Define $\displaystyle \pi (n)$ to be the primes $\displaystyle \leq n$. For a given $\displaystyle n$ (sufficiently large) let $\displaystyle p_1,p_2,...,p_{\pi (n)}$ be the primes $\displaystyle \leq n$.
Define $\displaystyle \Pi (n) = \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1}$.
Notice that $\displaystyle \left( 1 - \frac{1}{p_k} \right)^{-1} = \sum_{a_k =0}^{\infty} \frac{1}{p^{a_k}}$.
Therefore, we see that $\displaystyle \Pi (n) = \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1}$.
By the fundamental theorem of arithmetic any number $\displaystyle 1\leq m\leq n$ be be written as $\displaystyle m=p_1^{b_1}...p_{\pi (n)}^{b_{\pi(n)}}$ where $\displaystyle b_k\geq 0$.
It follows that,
$\displaystyle \sum_{k=1}^n \frac{1}{k} \leq \sum_{a_k \geq 0} (p_1^{a_1}p_2^{a_2}...p_{\pi(n)} ^{a_{\pi(n)}} )^{-1} = \Pi (n)$.
Since the harmonic series diverges it means $\displaystyle \lim ~ \Pi (n) = \infty \implies \lim ~ \log \Pi (n) = \infty$.
However,
$\displaystyle \log \Pi (n) = \log \prod_{k=1}^{\pi (n)} \left( 1 - \frac{1}{p_k} \right)^{-1} = - \sum_{k=1}^{\pi(n)} \log \left( 1 - \frac{1}{p_k} \right) = \sum_{k=1}^{\pi (n)} \sum_{j=1}^{\infty}\frac{1}{jp_k^j}$
Thus,
$\displaystyle \log \Pi (n) = \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right) + \sum_{k=1}^{\pi (n)} \sum_{j=2}^{\infty}\frac{1}{jp_k^j}$
Notice that,
$\displaystyle \sum_{j=2}^{\infty} \frac{1}{jp_k^j} \leq \sum_{j=2}^{\infty} \frac{1}{p_k^j} = \frac{1}{p_k^2(1-\tfrac{1}{p_k})} \leq \frac{2}{p_k^2}$
We have shown,
$\displaystyle \log \Pi (n) \leq \left( \frac{1}{p_1}+...+\frac{1}{p_{\pi(n)}} \right) + 2 \left( \frac{1}{p_1^2}+...+\frac{1}{p_{\pi(n)}^2} \right) \leq \sum_{k=1}^{\pi(n)} \frac{1}{p_k} + 2 \sum_{j=1}^{n} \frac{1}{j^2}$
Remember that $\displaystyle \sum_{j=1}^{\infty} \frac{1}{j^2} < \infty$.
What this means is that if $\displaystyle \sum_{k=1}^{\infty} \frac{1}{p_k} < \infty$ then $\displaystyle \log \Pi (n)$ would be bounded.
This is impossible, it is not bounded.
Thus, the sum of prime reciprocals must diverge.
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 $\displaystyle \subset$ is actually an equality.
although this doesn't change anything in the proof, but still the wikipedia author should have been more careful!
  |
