Given ${a_n}$ be the sequence of non zero integer with Following conditions Let ${a_n}$ be the sequence of non zero integer Satisfying 
$1).|a_n|<|a_{n+1}|\;\;\; \forall n$
$2) a_n \;\;\text{divides}\;\; a_{n+1}\;\; \forall n$
$3). \text{Every integer is a divisor of some } a_n$
Then $$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}\;\;\;\;\;\;\;.... is$$
$1).$ Absolutely  convergent and its sum is a rational number 
$2).$ Absolutely  convergent and its sum is a irrational number 
$3).$ Absolutely  convergent and its sum is a positive number 
$4).$ None of above
solution i tried-The given series is $$\sum_{n=1}^{\infty}\frac{1}{a_n}$$ 
so as per given condition $1$ by ratio test the given series converges Absolutly $$\left |\frac{ \frac{1}{a_{n+1}}}{\frac{1}{a_n}} \right |=\left | \frac{a_n}{a_{n+1}} \right |<1$$
and also it is sum of positive series so sum will be positive but,what can we say about its sum is it a rational or irrational ,i try to solve this question by taking different examples but no sequence satisfies all the given Conditions ,please provide me a hint 
Thank you
 A: You have already seen that points $1,3$ are false.
You have seen at least one series where the sum is irrational. While this is not immediately suggestive of $2$, you can still see if something goes wrong while assuming the sum is rational.

Suppose $\sum_{i=1}^\infty \frac 1{a_i} = \frac pq$ for some $p,q$ integers, $q > 0$.


*

*Let $N$ be the any index such that $q$ divides $a_N$ (by assumption, such an index exists). Then $q$ also divides each of $a_{N+1}, a_{N+2}$ etc. because $a_N$ is a divisor of each of these.

*Let $x_N = a_N\left(\frac pq - \sum_{i=1}^N \frac 1{a_i}\right)$. Why is $x_N$ an integer? (Hint : each denominator is a divisor of $a_N$). The previous bullet point tells us that if $x_N$ is an integer, so are $x_{N+1},x_{N+2}$ etc.

*Simultaneously, we have $x_N = \sum_{k=N+1}^\infty \frac{a_N}{a_k}$. We want to show that $|x_{N}| < 1$. Compare with the geometric series here : clearly $\left|\frac{a_N}{a_k}\right| \leq 2^{N-k}$ because of telescoping. Therefore, $|x_N| \leq \sum_{k=N+1}^\infty 2^{N-k} = 1$.

*But how do we rule out $|x_N| = 1$? That is simple : if $|x_N| = 1$, then each $\left|\frac{a_N}{a_k}\right|$ must be $2^{N-k}$, otherwise the inequality will become strict in the previous bullet point. This forces $|a_k| = 2^{k-N}|a_N|$ for every $k \geq N$. Ask yourself : if $p$ is an odd prime , $p>|a_N|$, then can any term of the sequence be a multiple of $p$? 

*Thus, we conclude that $|x_N| < 1$ for any $N$.

*Now , consider two consecutive $x_N$ and $x_{N+1}$. Their difference is $\frac{1}{a_i}$,which is non-zero. Therefore, at least one of $x_N$ or $x_{N+1}$ is non-zero. However, it also has magnitude smaller than $1$, contradicting the fact that it is an integer. 

*As a result of the contradiction, the sum must be irrational.
Compare with the proof of irrationality of $e$.
