1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.