Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $a_i$ be a sequence of 1's and 2's and $p_i$ the prime numbers.
And let $r=\displaystyle\sum_{i=1}^\infty p_i^{-a_i}$

Can r be rational, and can r be any rational > 1/2 or any real ?


ver.2:
Let k be a positive real number and let $a_i$ be 1 + (the i'th digit in the binary decimal expansion of k).

And let $r(k)=\displaystyle\sum_{n=1}^\infty n^{-a_n}$

Does $r(k)=x$ have a solution for every $x>\pi^2/6$, and how many ?

share|improve this question
2  
I have to wonder where these sums you have are coming from... –  J. M. Nov 6 '11 at 1:00
    
Eh, my answer was correct I believe (yes, any real in $[P(2),\infty)$) but the algorithm I gave was invalid. One algorithm would be to choose the $a_i$ one at a time depending on which stays under the desired $r$, and if ever both options are unavailable we simply go back and edit our last choice of $1$ to a $2$ and continue. I'm too tired right now to ... well, frankly, to figure out why I think this works. –  anon Nov 6 '11 at 1:23
4  
I would suggest a much more general question (there's really no reason to specialize to primes and restricted exponents). Let $a_i\le b_i$ for all $i=1,2,3,\dots$ Suppose $\sum a_i=A$ converges but $\sum b_i\to\infty$ diverges. Then the question would be how we prove that any number in $[A,\infty)$ can be represented by $\sum c_i$, where $c_i\in\{a_i,b_i\}$ for each $i$. –  anon Nov 6 '11 at 1:27
    
The only problem is that whenever you pick a $1$ instead of a $2$ you need to make sure that the tail of the sum with every prime given the power $-2$ still stays under the bound and you can't be too conservative with choosing $1$'s or you might never reach the limit. If this is easy to resolve, then the problem is simple. –  pki Nov 6 '11 at 1:28
3  
For shame, bad asker! Editing a question such that it becomes an entirely different one is rude and confusing; it makes all previous answers and comments incomprehensible. –  Henning Makholm Nov 6 '11 at 1:39

1 Answer 1

up vote 11 down vote accepted

The question with primes in the denominator:

The minimum that $r$ could possibly be is $C=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}$. However, a sequence of $1$s and $2$s can be chosen so that $r$ can be any real number not less than $C$. Since $\sum\limits_{i=1}^\infty\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)$ diverges, consider the sum $$ S_n=\sum_{i=1}^n b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right) $$ where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}>L-C$.

If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then $$ \sum\limits_{i=1}^\infty\frac{1}{p_i^{a_i}}=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}+\sum_{i=1}^\infty b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)=C+(L-C)=L $$ The question with non-negative integers in the denominator:

Changing $p_n$ from the $n^{th}$ prime to $n$ simply allows us to specify $C=\frac{\pi^2}{6}$. The rest of the procedure follows through unchanged. That is, choose any $L\ge C$ and let $$ S_n=\sum_{i=1}^n b_i\left(\frac{1}{i}-\frac{1}{i^2}\right) $$ where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}>L-C$.

If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then $$ \sum\limits_{n=1}^\infty\frac{1}{n^{a_i}}=\sum\limits_{n=1}^\infty\frac{1}{n^2}+\sum_{n=1}^\infty b_n\left(\frac{1}{n}-\frac{1}{n^2}\right)=C+(L-C)=L $$ We don't need to worry about an infinite final sequence of $1$s in the binary number since that would map to a divergent series.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.