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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!

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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

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