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

  • $\begingroup$ just want it's evaluation ? what has been tried ? $\endgroup$ – user451844 Sep 22 '17 at 23:24
  • $\begingroup$ So $a_n = \frac{\Gamma(21/9)}{\Gamma(10/9)} \cdot \frac{\Gamma(n+10/9)}{\Gamma(n+21/9)}$... I suppose since $\Gamma(n+k) \sim n^k \Gamma(n)$ as $n \to \infty$ for $k$ an integer, it would be reasonable to suppose Stirling's approximation will give $a_n = \Theta(n^{-10/9})$ which would at least convince me it converges... $\endgroup$ – Daniel Schepler Sep 22 '17 at 23:29
  • $\begingroup$ @DanielSchepler - Thanks for your observation! $\endgroup$ – hypergeometric Sep 22 '17 at 23:34
  • $\begingroup$ It appears that the sum is $0.909090909\dots = \frac{10}{11}$. $\endgroup$ – orlp Sep 22 '17 at 23:38
  • $\begingroup$ @orlp - Yes, that's correct. $\frac {10}{11}$. The question is why :) $\endgroup$ – hypergeometric Sep 22 '17 at 23:41

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 $20 a_1 = \frac{10}{11}$.

  • $\begingroup$ Might be clearer to write the last summation as $$\sum_{n=1}^\infty (9n+11)a_n-(9(n+1)+11)a_{n+1}$$. $\endgroup$ – hypergeometric Sep 24 '17 at 3:08
  • $\begingroup$ Nice answer! (+1). Accepted! $\endgroup$ – hypergeometric Sep 24 '17 at 3:09

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