# What can be said about the sum of the series?

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.

• 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 $a_n\sim \pm n!$? Aug 15, 2020 at 17:03
• @DanieleTampieri Well, one could also choose $a_n=(2n)!$, but then $a_n\not\sim \pm n!$
– Zuy
Aug 15, 2020 at 17:12
• 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?
– lulu
Aug 15, 2020 at 17:16
• @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. Aug 15, 2020 at 18:53
• @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.
– lulu
Aug 15, 2020 at 19:03

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.

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