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Let $\{a_n \}_{n \geq 1}$ be a sequence of non-zero integers satisfying

I. $|a_n| \lt |a_{n+1}|,$ for all $ n \geq 1$

II. $a_n$ divides $a_{n+1},$ for all $n \geq 1$ and

III. every integer is a divisor of some $a_n.$

Then $\displaystyle\sum\limits_{n=1}^{\infty} \frac {1} {a_n}$ is

(a) absolutely convergent and it's sum is a rational number.

(b) absolutely convergent and it's sum is an irrational number.

(c) absolutely convergent and it's sum is a positive number.

(d) none of the above.

My attempt $:$ It is easy to see that $\displaystyle\sum\limits_{n=1}^{\infty} \frac {1} {a_n}$ is absolutely convergent. Let $a_{k+1} = m_k\ a_{k},$ for $k \geq 1.$ By (I) it follows that $|m_k| \geq 2,$ for all $k \geq 1.$ So we have

\begin{align*} \sum\limits_{n=1}^{\infty} \frac {1} {|a_n|} & = \frac {1} {|a_1|} + \frac {1} {|a_2|} + \frac {1} {|a_3|} + \cdots \\ & = \frac {1} {|a_1|} + \frac {1} {|m_1|\ |a_1|} + \frac {1} {|m_2|\ |a_2|} + \cdots \\ & = \frac {1} {|a_1|} + \frac {1} {|m_1|\ |a_1|} + \frac {1} {|m_1|\ |m_2|\ |a_2|} + \cdots \\ & \leq \frac {1} {|a_1|} \left ( 1 + \frac {1} {2} + \frac {1} {2^2} + \cdots \right ) \\ & = \frac {2} {|a_1|} < \infty \end{align*}

So $\displaystyle\sum\limits_{n=1}^{\infty} \frac {1} {a_n}$ is absolutely convergent. Clearly (a) is false because we can take $a_n = n!,$ for all $n \geq 1.$ Then the sum is $e-1,$ which is clearly irrational. I may as well take $a_n = -n!,$ for all $n \geq 1.$ Which makes the sum $1-e,$ a negative quantity. Hence (c) is also false. But how can I conclude that whether the sum is always irrational or not? Any help in this regard will be highly appreciated.

Thanks in advance.

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  • $\begingroup$ It seems to me that the key data to exploit for the solution of the problem is hypothesis III: could it be that this leads to an expression like $a_n\sim \pm n!$? $\endgroup$ Aug 15, 2020 at 17:03
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    $\begingroup$ @DanieleTampieri Well, one could also choose $a_n=(2n)!$, but then $a_n\not\sim \pm n!$ $\endgroup$
    – Zuy
    Aug 15, 2020 at 17:12
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    $\begingroup$ Here's a simpler question I, for one, can't answer: Say you take the usual series $\sum \frac 1{n!}$ and change the signs of whichever terms you like. The resultant series always converges, of course...is it always transcendental? Is it always irrational? $\endgroup$
    – lulu
    Aug 15, 2020 at 17:16
  • $\begingroup$ @lulu For $\sum \frac{\pm 1}{n!}$ the sum is always irrational, essentially the same proof as for the irrationality of $e$ works. I don't know about transcendentality. $\endgroup$ Aug 15, 2020 at 18:53
  • $\begingroup$ @DanielFischer Yes, you're right about irrationality. I'd hoped that this result would quickly generalize into a statement about the OP's $\sum \frac 1{a_n}$. But I can't see that it does. Even though the $a_n$ share some properties with $n!$ I don't see how to force the same proof through. $\endgroup$
    – lulu
    Aug 15, 2020 at 19:03

1 Answer 1

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Let $$ R_m = \sum_{n > m} \frac 1{a_n} .$$

Lemma: $0 < \displaystyle |R_m| < \frac1{|a_m|} .$

Proof: Let $r \ne 2$ be a prime number that is not a factor of $a_{m}$. Then there exists $m' > m$ such that $r | a_{m'}$. Thus for $n > m$ we have $|a_n| \ge 2^{n-m} |a_m|$, and for $n \ge m'$ we have $|a_n| \ge r 2^{n-m-1} |a_m|$. Thus $$|R_m| \le \sum_{n > m} \left|\frac {1}{a_n} \right| \le \frac1{|a_{m}|}\left(\sum_{n=m+1}^{m'-1} 2^{m-n} + \frac 2 r \sum_{n=m'}^\infty 2^{m-n}\right) < \frac1{|a_m|}.$$ For the lower bound, $R_m = \frac1{a_m} + R_{m+1}$, so $$|R_m| = \frac1{|a_m|} - |R_{m+1}| \ge \frac1{|a_m|} - \frac1{|a_{m+1}|} > 0 .$$

$\square$

Suppose $$S = \sum_n \frac1{a_n} = \frac pq $$ where $p,q \ne 0$ are integers.

For some $m$, we have $q | a_m$. Then $a_m S = \text{integer} + a_m R_m$ is an integer. From the Lemma, we see that $0 < |a_m R_m| < 1$. So $a_m R_m$ cannot be an integer, and so $S$ cannot be rational.

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  • $\begingroup$ It's kind of similar to the proof $e$ is irrational. Very nice answer +1. $\endgroup$ Aug 15, 2020 at 19:59

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