This is the sum,

$$e'=\sum_{n=1}^\infty \frac{1}{F_{n}!}$$ where $F_{n}$ is the $n^{th}$ Fibonacci number.

Is it possible to prove that it will converge to a transcendental number?

Edit: Proof of irrationality-:

But first some lemmas,

$F_q\geq q \,\,\,\,$ $\forall q\geq5$

If $P_n$ represents the partial sums of $e'$, then $e'>P_n$ $\forall n\in\mathbb N$

If $e'=\frac{a}{b}$ for some $a,b\in\mathbb N$ , then $b>4$.

Reason: Obviously $b=1$ is not possible. If $b=2$, then $a=5$ or $a=6$ are the only possibilities. However, in both we get $e'=2.5$ and $e'=3$ respectively, which is again not possible, since adding after 3 terms makes $e'>2.5$ and $e'<3$ because $e'<e$. Similarly, we can also prove that $b\neq 3$ and $b\neq4$.

$\therefore b\geq5$

Now we just need one more inequality before the main proof. Namely $e'-P_q<\frac{1}{F_q!} \forall q>4$

Reason: Recall that $$\sum_{k=n+1}^{\infty}\frac{1}{k!}<\frac{1}{n!}\;\;\;\forall n\geq1$$

also $e'-P_q=\frac{1}{F_{q+1}!}+\frac{1}{F_{q+2}!}+\cdots$

$\therefore$ from (1), $$e'-P_q<\frac{1}{F_{q}!}\;\;\;\forall q\geq5$$ (Actually this is true for all $q\geq2$

Now, the main proof.

*Theorem.* $e'$ is irrational

*Proof.* Assume $\exists p,q \; \in \mathbb N$ such that $e'=\frac{p}{q}$, where $gcd(p,q)=1$

then $q\geq5$ and $\frac{p}{q}-P_q<\frac{1}{F_{q}!}$

Now, $\left(\frac{p}{q}-P_q\right)=\left(\frac{p}{q}-P_q\right)\frac{F_{q}!}{F_{q}!}$

=$\frac{pF_{q}!}{qF_{q}!}-\frac{F_{q}!}{F_{q}!}-\frac{F_{q}!}{F_{q}!}-\frac{1}{2!}\frac{F_{q}!}{F_{q}!}-\frac{1}{3!}\frac{F_{q}!}{F_{q}!}-\cdots-\frac{1}{F_{q-1}!}\frac{F_{q}!}{F_{q}!}-\frac{1}{F_{q}!}\frac{F_{q}!}{F_{q}!}$

=$\frac{C}{F_{q}!}$

where $C\in\mathbb Z$

however, from inequality (4), we know that $\frac{C}{F_{q}!}<\frac{1}{F_{q}!}$, which implies that $C\leq0$,

but then $\frac{p}{q}-P_q\leq0$, which contradicts (2).

hence, by contradiction $e'$ is irrational.

(Please let me know if there is any mistake)