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Conjecture:

There is no $b,\{a_n\}_{n=1}^{\infty}$ such that $b,a_n \in \mathbb{N}^+, a_{n+1}\ge a_n$, $$\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\infty\qquad\text{and}\qquad\frac{1}{b}= \sum_{n=1}^{\infty}\frac{1}{a_{n}}.$$ This is just my guess, and it would be nice if someone could give a counter example.

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  • $\begingroup$ @DanieleTampieri I don't think $\frac{1}{e-1}\in\mathbb{N}^+$. $\endgroup$
    – gmvh
    Jun 7 at 7:35
  • $\begingroup$ @gmvh you're right. Thanks for pointing it out. $\endgroup$ Jun 7 at 7:36
  • $\begingroup$ Can you give an example of an $a_n$ that satisfies the given condition? I only see something like $2^{2^n}$ which is not something one encounter on a daily basis $\endgroup$
    – LegNaiB
    Jun 7 at 7:56
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    $\begingroup$ @LegNaiB You gave an example yourself I think. "$a_n$ must be something encountered on a daily basis" is not among the hypotheses. :) $\endgroup$ Jun 7 at 7:57
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    $\begingroup$ This is not that an exotic condition, even in the context of such sums. For instance, for integer $q\ge 1$ and $a_n=q^nn!$, the sum is $\exp(1/q)$, which is irrational. $\endgroup$
    – YCor
    Jun 7 at 8:02
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Actually, this happens for all natural $b$. Notice that $$ \frac{1}{b}=\frac{1}{b+1}+\frac{1}{b^2+b} $$ and iterate this identity. You will get $$ \frac{1}{b}=\frac{1}{b+1}+\frac{1}{b^2+b+1}+\frac{1}{(b^2+b)(b^2+b+1)+1}+\ldots, $$ i.e. $$ \frac{1}{b}=\sum_n \frac{1}{a_n}, $$ where $a_1=b+1$ and $a_{n+1}=a_n^2-a_n+1$. This is similar to Sylvester's sequence and one can easily see that this sequence grows doubly exponentially.

EDIT: To produce example for which $a_{n+1}/a_n$ goes to infinity as slowly as you want, use the identity $$ \frac{1}{q-1}=\frac{1}{q}+\ldots+\frac{1}{q^n}+\frac{1}{q^n(q-1)}. $$ This identity implies, for example, that if you got some finite sum like $$ 1/b=1/a_1+\ldots+1/a_k $$ and your last term is divisible by $q-1$, you can insert (essentially) a finite geometric progression instead of the last term. For example, start with $$ 1=1/2+1/4+1/4. $$ Replace $1/4$ with $$ 1/6+1/18+1/54+1/108. $$ (here we take $q=3$ and $n=3$). Then for $1/108$ you can take $q=n=4$ etc.

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    $\begingroup$ Thank you very much for your accurate and concise answer.And even more, if we add one more condition$$\lim_{n\rightarrow \infty}\frac{a_{n} a_{n+2}}{a_{n+1}^2}=1$$,are there another way to construct an equation like this answer? $\endgroup$
    – LMZ
    Jun 7 at 10:24
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    $\begingroup$ @LMZ yeah, you can construct examples with $\ln a_n/n$ going to $\infty$ as slowly as you want (and as smoothly as you want, I guess), see my edit $\endgroup$ Jun 7 at 10:59

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