<s>[INST] Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:

If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
  $$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} = \frac{(a_{1} + 2) (a_{2} + 2) \cdots (a_{n-1} + 2)}{2^{n}},
$$
  find $ \displaystyle \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $.

 [/INST] Since $a_1=1$ and
$$
a_{n+1}=\prod_{k=1}^n\left(1+\frac{a_k}2\right)\tag{1}
$$
we have that
$$
a_n\ge1\tag{2}
$$
and
$$
a_{n+1}=a_n\left(1+\frac{a_n}2\right)\tag{3}
$$
Subtracting $a_n$ from both sides of $(3)$ gives
$$
a_{n+1}-a_n=\frac{a_n^2}2\tag{4}
$$
and since $a_1=1$, $(1)$ and $(2)$ show that
$$
a_n\ge\left(\frac32\right)^{n-1}\tag{5}
$$
Using $(3)$ and then $(4)$ yields
$$
\begin{align}
\frac1{a_n+2}
&=\frac12\frac{a_n}{a_{n+1}}\\
&=\frac{a_{n+1}-a_n}{a_{n+1}a_n}\\
&=\frac1{a_n}-\frac1{a_{n+1}}\tag{6}
\end{align}
$$
Summing $(6)$ yields
$$
\begin{align}
\sum_{k=1}^n\frac1{a_k+2}
&=\sum_{k=1}^n\left(\frac1{a_k}-\frac1{a_{k+1}}\right)\\
&=\frac1{a_1}-\frac1{a_{n+1}}\tag{7}
\end{align}
$$
Taking the limit of $(7)$ and applying $(5)$ gives
$$
\begin{align}
\sum_{k=1}^\infty\frac1{a_k+2}
&=\frac1{a_1}\\
&=1\tag{8}
\end{align}
$$

Original Sequence
It has been pointed out that the recursion was and is again different than when I wrote my answer. We have
$$
a_1=1, a_2=\frac34, a_3=\frac{33}{32}
$$
$(3)$ and $(4)$ still hold, but since $(2)$ only holds for $n\ge3$, $(5)$ should read
$$
a_n\ge\left(\frac32\right)^{n-3}
$$
and $(6)$ only holds for $n\ge2$. Furthermore, $(7)$ should read
$$
\begin{align}
\sum_{k=2}^n\frac1{a_k+2}
&=\sum_{k=2}^n\left(\frac1{a_k}-\frac1{a_{k+1}}\right)\\
&=\frac1{a_2}-\frac1{a_{n+1}}
\end{align}
$$
and $(8)$ should read
$$
\begin{align}
\frac1{a_1+2}+\sum_{k=2}^\infty\frac1{a_k+2}
&=\frac1{a_1+2}+\frac1{a_2}\\
&=\frac53
\end{align}
$$
As it is in JimmyK4542's answer.
</s>