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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? Dbfirs 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 $\pi =\underset{n\to \mathrm{\infty }}{\lim }{a}_n$. 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 Template: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 $f^{\prime }(0)=0$. Define the Newton iteration by $x_0=1$, $x_{n+1}=x_n-f(x_n)/f^{\prime }(x_n)$. Then $\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\pi$. 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 $1\le s,t\le n$ and $\gcd (s,t)=1.$ Then $\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{6n^2}{a_n}}=\pi .$ --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 $3\frac{10}{71}<\pi <3\frac{1}{7}$ (→ Measurement of a Circle#Proposition three). --CiaPan (talk) 07:04, 29 May 2017 (UTC)

Consider also continued fraction presented in Pi#Continued fractions – consecutive truncations of any of them result in a sequence of rational numbers Template:Nowrap. --CiaPan (talk) 09:22, 29 May 2017 (UTC)