Under a condition, $\frac{1}{b } = \sum_{n=1}^{\infty}\frac{1}{a_{n}}$ will never happen

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.

• @DanieleTampieri I don't think $\frac{1}{e-1}\in\mathbb{N}^+$.
– gmvh
Jun 7, 2021 at 7:35
• @gmvh you're right. Thanks for pointing it out. Jun 7, 2021 at 7:36
• 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 Jun 7, 2021 at 7:56
• @LegNaiB You gave an example yourself I think. "$a_n$ must be something encountered on a daily basis" is not among the hypotheses. :) Jun 7, 2021 at 7:57
• 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.
– YCor
Jun 7, 2021 at 8:02

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.
• 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?
• @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 Jun 7, 2021 at 10:59