Wikipedia:Reference desk/Archives/Mathematics/2017 May 28

= May 28 =

Sequence with π as limit
Can a sequence having the number π as limit be constructed?(Thanks)--82.79.115.90 (talk) 11:23, 28 May 2017 (UTC)


 * I assume you mean a sequence that doesn't implicitly involve pi in each term?   D b f i r s   11:26, 28 May 2017 (UTC)


 * There are many examples. One, due to Leibniz, is $$a_n = 4\sum_{k=0}^n \frac{(-1)^k}{2k+1}$$.  Then $$\pi=\lim_{n\to\infty}a_n$$.  Sławomir Biały  (talk) 11:29, 28 May 2017 (UTC)


 * I wonder, by taking into consideration the answer, if there is a sequence (with non-implicit reference to pi) which is also NOT a series that satisfies the initial question this time with this strict specification explicitly stated?--82.79.115.90 (talk) 13:57, 28 May 2017 (UTC)


 * I object slightly that the question is now ill-posed, since every sequence can be written as the sequence of partial sums of a series (of differences) and vice-versa.  But in any case, if the question is more along the lines of "is there a sequence, that is naturally a sequence and not a series, that converges to $\pi$", then the answer is still yes.  Let $$f(x)$$ denote the unique solution to the second order differential equation $$f''(x)+f(x)/4=0$$ satisfying $$f(0)=1$$ and $$f'(0)=0$$.  Define the Newton iteration by $$x_0=1$$, $$x_{n+1}=x_n-f(x_n)/f'(x_n)$$.  Then $$\lim_{n\to\infty}x_n=\pi$$.   Sławomir Biały  (talk) 14:29, 28 May 2017 (UTC)


 * Does the strict sequence associated to the Leibniz series have the same limit with the series? (Probably not, I guess, given the alternating nature of the deSeriesified Leibniz sequence.)--82.79.115.90 (talk) 14:01, 28 May 2017 (UTC)


 * I don't know what you mean. The Leibniz series is conditionally convergent.   Sławomir Biały  (talk) 14:29, 28 May 2017 (UTC)
 * There's a bunch listed here. Abductive  (reasoning) 17:58, 28 May 2017 (UTC)


 * How about $$a_n = 2 \arctan n$$ ...? --CiaPan (talk) 18:00, 28 May 2017 (UTC)


 * Another one: let $$a_n$$ be the number of pairs of integers $$(s,t)$$ with $$1 \leq s,t \leq n$$ and $$\gcd(s,t)

= 1.$$ Then $$\lim_{n\to\infty} \sqrt{\frac{6n^2}{a_n}} = \pi.$$ --Deacon Vorbis (talk) 20:22, 28 May 2017 (UTC)


 * Possibly one of the first to invent (discover?) such sequence was Archimedes of Syracuse who used a few initial terms to find the famous approximation $$ 3\tfrac{10}{71} < \pi < 3\tfrac 17$$ (→ Measurement of a Circle). --CiaPan (talk) 07:04, 29 May 2017 (UTC)


 * Consider also continued fraction presented in Pi – consecutive truncations of any of them result in a sequence of rational numbers converging to &pi;. --CiaPan (talk) 09:22, 29 May 2017 (UTC)