Summation of Products $\sum\limits_{n=1}^\infty \frac 1{11}\cdot \frac {10}{20}\cdot \frac {19}{29}\cdots \frac{9n+1}{9n+11}$ This is an extension of a nice question posted recently on MSE.
One of the solutions posted was very interesting and this is an attempt to  extend it to a general case.

Evaluate 
  $$\sum_{n=1}^\infty a_n,\qquad a_n=\frac 1{11}\cdot \frac {10}{20}\cdot \frac {19}{29}\cdot \frac{28}{38}\cdots \frac{9n+1}{9n+11}$$
  $$a_n = \prod_{k=0}^n\frac{9k+1}{9k+11}$$

 A: In general, consider:
$$\sum_{n=1}^\infty \frac{\Gamma(n+a)}{\Gamma(n+a+k)} = \frac{1}{k-1} \sum_{n=1}^\infty [(n+a+k-1) - (n+a)] \cdot \frac{\Gamma(n+a)}{\Gamma(n+a+k)} = \\ \frac{1}{k-1} \sum_{n=1}^\infty \left( \frac{\Gamma(n+a)}{\Gamma(n+a+k-1)} - \frac{\Gamma(n+a+1)}{\Gamma(n+a+k)} \right).$$
If $k > 1$, then this is a telescoping series for which the term $\frac{\Gamma(n+a)}{\Gamma(n+a+k-1)} \to 0$ as $n \to \infty$.  Therefore, the sum is equal to $\frac{1}{k-1} \cdot \frac{\Gamma(a+1)}{\Gamma(a+k)} = \frac{\Gamma(a+1)}{(k-1) \Gamma(a+k)}$.
Now, the originally desired sum is equal to
$$\sum_{n=1}^\infty \frac{\Gamma(11/9)}{\Gamma(1/9)} \cdot \frac{\Gamma(n+10/9)}{\Gamma(n+20/9)} = \frac{\Gamma(11/9)}{\Gamma(1/9)} \cdot \frac{\Gamma(19/9)}{(10/9-1) \Gamma(20/9)} = \frac{10}{11}.$$
Note: Expanding the argument a bit, this is equivalent to observing that
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty [(9n+11)a_n - (9n+10)a_n] = \sum_{n=1}^\infty [(9n+11)a_n - (9n+20)a_{n+1}]$$
which again telescopes, so the sum is equal to $20 a_1 = \frac{10}{11}$.
