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.



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


