Inequality With Recurrent Relation $a_{1}=1$, $~$ $a_{n+1}-a_{n}=\sqrt{\dfrac{a_{n+1}^{2}-1}{2}}+\sqrt{\dfrac{a_{n}^{2}-1}{2}}$ , $~$ $a_{n+1}>a_{n}$
Prove that $~$ $\displaystyle\sum_{k=1}^{\infty}\frac{1}{a_{n}}<e$
 A: Note that sequence $\{a_n, ~ n\in \mathbb{N}\}$ is non-decreasing (because RHS can't be negative, if $a_n\in \mathbb{R}$).
First, let's consider increasing sequence $\{a_n, ~ n\in \mathbb{N}\}$:
$$
a_{n+1}>a_n.\tag{1}
$$
1.
Let's show that there is other (easy) recurrent relation
$$
a_n = 6a_{n-1}-a_{n-2}, \qquad (n\geqslant 3),\tag{2}
$$ 
where $a_1=1,a_2=3$.
So, we'll have integer sequence
$$
1,~~ 3,~~ 17,~~ 99,~~ 577,~~ 3363,~~ 19601,~~ \ldots\tag{3}
$$
It is easy to see that $a_2$ can be $1$ either $3$ (as solution of quadratic equation).
We'll consider case $a_2=3$ here, because case $a_2=1$ doesn't hold ineq. $(1)$.
Denote $b_n = \sqrt{\dfrac{a_n^2-1}{2}}$ . Then 
$$
a_{n+1}-a_n = b_{n+1}+b_n,\qquad (n\in \mathbb{N}),\tag{4}
$$
and also $a_{n+1}-a_n = \sqrt{\dfrac{a_{n+1}^2-1}{2}}+b_n$.
Now we'll find $a_{n+1}$ as expression of $a_n, b_n$.
Denote $\color{purple}{x} = a_{n+1}$.
$$\color{purple}{x}-a_n = \sqrt{\dfrac{\color{purple}{x}^2-1}{2}} + b_n;$$
$$\color{purple}{x}-(a_n+b_n) = \sqrt{\dfrac{\color{purple}{x}^2-1}{2}}; ~~~~~ (\color{purple}{x}\geqslant a_n+b_n);\tag{5}$$
$$2\Bigl(\color{purple}{x}^2-2(a_n+b_n)\color{purple}{x} +(a_n+b_n)^2\Bigr)= \color{purple}{x}^2-1.$$
$$\color{purple}{x}^2-4(a_n+b_n)\color{purple}{x} +2(a_n+b_n)^2+1= 0.$$
Solving this quadratic equation, we find $2$ solutions:
$a_{n+1} = \color{purple}{x} = a_n$; (but this value doesn't hold ineq. $(1)$, and holds $(5)$ when $b_n=0$ only);
$a_{n+1} = \color{purple}{x} =3a_n+4b_n$.
We will focus on the last expression:
$$a_{n+1}  = 3a_n + 4b_n, \qquad (n\in \mathbb{N}).\tag{6}$$
$(4),(6)\implies$ 
$$b_{n+1} = a_{n+1}-(a_n+b_n) = 2a_n+3b_n, \qquad (n\in \mathbb{N}).\tag{7}$$
$(6),(7)\implies$
$$a_{n+1} = 3(3a_{n-1} + 4b_{n-1}) +4(2a_{n-1}+3b_{n-1}) = 17a_{n-1} + 24b_{n-1}, \qquad (n\geqslant 2);$$
$$a_{n+1} = (18a_{n-1} + 24b_{n-1}) -a_{n-1} = 6(3a_{n-1} + 4b_{n-1}) -a_{n-1} = 6a_n-a_{n-1},\qquad (n\geqslant 2).$$
Statement $(2)$ is proved.
2.
Since $a_{n-2}<a_{n-1}$ (see $(1)$), then 
$$
a_n > 5a_{n-1}, \qquad (n\geqslant 3),
$$
other words
$$
a_n > 3 \cdot 5^{n-2}, \qquad (n\geqslant 3).\tag{8}
$$
3.
Then 
$$
\sum_{n=1}^\infty \dfrac{1}{a_n} < 1+\dfrac{1}{3} + \sum_{n=3}^\infty \dfrac{1}{3\cdot 5^{n-2}} = 1+\dfrac{1}{3} + \dfrac{1}{12} = \dfrac{17}{12} <2<e.\tag{9}
$$
Bound $\dfrac{17}{12} = 1+\dfrac{5}{12}$ is more sharp estimation than $e$.

I missed the case, when can be $a_{n+1} = a_n$ (non-decreasing sequence).
It can be when $b_{n+1}=b_n=0$ (see $(4)$), other words $a_{n+1}=a_n=1$.
So, if we will consider $a_2=a_1$, then, as Don Antonio noted in comments,
it generates constant sequence $1,1,1,1,\ldots$.
For this sequence we will have
$\sum\limits_{n=1}^\infty \dfrac{1}{a_n} = \sum\limits_{n=1}^\infty 1~~$ (divergent series).
