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} $.

 A: 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.
A: For $n \ge 2$, we have: $a_{n+1} = \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)(a_n+2)}{2^{n+1}}$ $= \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)}{2^n}\cdot \dfrac{a_n+2}{2} = \dfrac{a_n(a_n+2)}{2} = \dfrac{1}{2}a_n^2+a_n$. 
Thus, $\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}} = \dfrac{a_{n+1}-a_n}{a_na_{n+1}} = \dfrac{\tfrac{1}{2}a_n^2}{a_n \cdot \tfrac{1}{2}a_n(a_n+2)} = \dfrac{1}{a_n+2}$. 
Therefore, $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{a_n+2} = \dfrac{1}{a_1+2} + \sum_{n=2}^{\infty}\left[\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}\right] = \dfrac{1}{a_1+2} + \dfrac{1}{a_2} = \dfrac{1}{1+2} + \dfrac{1}{\tfrac{3}{4}} = \boxed{\dfrac{5}{3}}$, 
where we have used the fact that the sum is telescoping and $a_n \to \infty$ as $n \to \infty$.

EDIT: We can show that $a_n \to \infty$ without assuming that the series converges. 
Note that $\dfrac{a_{n+1}+1}{2} = \dfrac{1}{4}a_n^2+\dfrac{1}{2}a_n+\dfrac{1}{2} = \left(\dfrac{a_n+1}{2}\right)^2+\dfrac{1}{4} \ge \left(\dfrac{a_n+1}{2}\right)^2$ for all $n \ge 2$. 
Since $a_3 = \dfrac{33}{32}$, by induction we have $\dfrac{a_n+1}{2} \ge \left(\dfrac{a_3+1}{2}\right)^{2^{n-3}} = \left(\dfrac{65}{64}\right)^{2^{n-3}}$ for all $n \ge 3$. 
Clearly, $\displaystyle\lim_{n\to\infty}\left(\dfrac{65}{64}\right)^{2^{n-3}} = +\infty$. Therefore, $\displaystyle\lim_{n\to\infty}a_n = +\infty$ as desired. 
