My task was to prove that the limit in the title exists and to calculate it. First I showed that the sequence is monotonically decreasing $\left(\frac{a_{n+1}}{a_{n}} = \frac{n+1}{\frac{2}{7} + n + 1} < 1\right)$ and it's obvious that $a_{n} > 0$ for all $n$. Therefore, by the Monotone Convergence theorem $(a_n)_{n \geq 1}$ must converge.
My first idea was to use the relation I obtained earlier between $a_{n+1}$ and$a_{n}$: $a_{n+1} = \frac{n+1}{\frac{2}{7} + n + 1} \cdot a_{n}$ and passing the limit on both sides, but I arrive at the trivial equality that $\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} a_{n}$.
My second idea was to rationalize the denominator to see if a more familiar expression will arise, here is what I arrived at:$$a_{n} = \frac{7^n \cdot n!}{(2+7 \cdot 1)(2 + 7 \cdot 2)...(2 + 7 \cdot n)}$$ which didn't help.
If it is of any help, the first exercise was to study the convergence of: $$\sum_{n=1}^{\infty} \frac{n!}{(2x+1)...(2x+n)} \text{ for } x \in (0, +\infty)$$ I determined that it converges when $x \in \left(\frac{1}{2}, \infty\right)$ by Raabe-Duhammel and diverges otherwise.
