I am getting crazy on this series! I found this in a handwritten old book without a reference. I could not figure out how it is built but the series numerically seems to converge to $\pi$. \begin{align} a_1&=\frac{16}{3}\\ a_2&=\frac{56}{15}\\ a_3&=\frac{362}{105}\\ a_4&=\frac{1051}{315}\\ a_5&=\frac{90913}{27720}\\ a_6&=\frac{2339483}{720720}\\ a_7&=\frac{9294869}{2882880}\\ a_8&=\frac{314539061}{98017920}\\ a_9&=\frac{95291361359}{29797447680}\\ a_{10}&=\frac{27155335099}{8513556480}\\ a_{11}&=\frac{2493237983453}{783247196160}\\ a_{12}&=\frac{24892232679053}{7832471961600}\\ a_{13}&=\frac{596632945162997}{187979327078400}\\ a_{14}&=\frac{34567420288501151}{10902800970547200}\\ a_{15}&=\frac{4282497882211187099}{1351947320347852800}\\ a_{16}&=\frac{8558465078579558323}{2703894640695705600}\\ ...\\ a_{\infty}&=\pi \end{align}

I have observed that the denominators include multiplication of odd numbers while $2^j$ is also always around. Sometimes the odd numbers appear in a row sometimes they are not in order. For the numerator I do not see much of a pattern!

  • $\begingroup$ What was the context around this sequence? Was the book covering Taylor series, for example? The sequence is unknown to me, but it seems to converge rather slowly, as even $a_{16}$ is only around $4.165$, so it's $0.02$ off... $\endgroup$
    – 5xum
    Sep 3, 2015 at 8:56
  • $\begingroup$ @5xum The context around the sequence is series expansions ... I have thought of different possible functions that could have double factorial in the denominator but no success so far. $\endgroup$
    – Math-fun
    Sep 3, 2015 at 9:01
  • $\begingroup$ The numerator does not come up in the OEIS: oeis.org/… Nor does the denominator: oeis.org/… $\endgroup$
    – Hrodelbert
    Sep 3, 2015 at 9:06
  • 1
    $\begingroup$ is it somehow related to the Wallis product? en.wikipedia.org/wiki/Wallis_product $\endgroup$ Sep 3, 2015 at 9:07
  • $\begingroup$ @DavidQuinn many thanks for the nice hint. :-) $\endgroup$
    – Math-fun
    Sep 3, 2015 at 11:45

1 Answer 1


I can give a recursion which seems to fit at least the early values:

$$a_n = a_{n-1} - \dfrac{(2n-4)!}{(n-2)!\,(n-1)!\,(2n+1)\,2^{2n-7}}$$ with a suggestion of Catalan numbers or double factorials in there.

As far as I can tell this gives $a_{513} \approx 3.141722$ so it may well be converging on $\pi$ from above.

  • $\begingroup$ This looks very nice! Many thanks! $\endgroup$
    – Math-fun
    Sep 3, 2015 at 12:09

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.