prove that the infinite sum - $∑(1/2)^p$, where $p$ runs over all the $prime$ numbers,is $irrational$

one idea that may work is to use the given lemma

$Lemma:$ $α$ is an irrational number iff there exists two convergent integer sequences ${a_n}$ and $b_n$ such that $(a_n-αb_n)≠0$ for all $n$. but $lim(a_n-αb_n)=0$

The proof is just by contradiction by assuming $α$ to be rational.I have tried to work out this lemma but failed .Plz try this.

  • 1
    $\begingroup$ A different hint, that can be formalized: an irrational number has a non-repeating expansion in any base. What does the binary expansion of your infinite sum look like? Can it repeat? $\endgroup$ – Steven Stadnicki Oct 12 '17 at 17:51
  • $\begingroup$ There is something wrong in your lemma. $\alpha$ is rational iff there exists integers $a,b$ such that $a - \alpha b = 0$. $\endgroup$ – reuns Oct 12 '17 at 17:53
  • $\begingroup$ The lemma is actually correct, but it's a strange one for this problem - it says that irrational numbers can be 'approximated' better than rational numbers can; if $\alpha$ is rational then there's a constant $C$ such that for all $\frac ab\neq\alpha$, then $\left|\frac ab-\alpha\right|\geq \frac Cb$. (Contrast this with irrational numbers, where we have a constant $D$ and infinitely many $a,b$ with $\left|\frac ab-\alpha\right| \leq \frac D{b^2}$. ) $\endgroup$ – Steven Stadnicki Oct 12 '17 at 22:07

Hint: look at the similar number $$\sum_{i = 1}^\infty \left(\frac{1}{10}\right)^{p_i} \approx 0.0110101000101000101$$ (be sure to notice that's "approximately," not "equals").

Clearly the $1$s will occur in the prime positions. You should already know that there are infinitely many primes. You should also know that their distribution feels kind of random. With one exception, any two $1$s are separated by an odd number of $0$s. But other than that... kind of random.

If this number was rational, we could find integers $m$ and $n$ such that one divided by the other gives this number. Try truncating the number at the prime positions. That does give rational numbers, but... $$\frac{1}{100}, \frac{11}{1000}, \frac{1101}{100000}, \frac{110101}{10000000}, \ldots$$


As in base $10$, rational numbers written in base $2$ are either terminating or, eventually, periodic. This sum is neither.


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