Is there a positive series of rational numbers that converge to $\pi$?

Question

I'm looking for a series of rational numbers that approach $\pi$ that are all non-negative. In addition, I also want this series to have a general rule. So something like $\dfrac31+\dfrac1{10}+\dfrac4{100}+\ldots$ is not gonna cut it.

Does such a series exist?

$4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}+\dots\pm\frac{4}{2k+1}\pm\dots = \pi$. You specify "all non-negative" so just group adjacent terms together. — JMoravitz, Dec 28 '20 at 00:46
For all we know, there is a general rule giving $3+0.1+0.04+\cdots$. — Gerry Myerson, Dec 28 '20 at 01:04
I mean, you can go with $$\pi=\sum_{k=0}^\infty 10^{-k}\left\lfloor 10^k\pi\right\rfloor-10^{-k+1}\left\lfloor 10^{k-1}\pi\right\rfloor$$ though apparently you don't like it. — , Dec 28 '20 at 01:10
The existence of such a series is certain for every positive real number. Not always easy (or even possible) is to find a sum with a closed form general term doing the job. A large class of real numbers including $\pi$ however allows such a representation. — Peter, Dec 28 '20 at 09:44

score 4 · Answer 1 · answered Dec 28 '20 at 09:15

This one is not bad $$\pi=\sum_{n=0}^\infty \frac{ (2 n)\text{!!}}{2^{n-1}\,(2 n+1)\text{!!}}$$ The terms form the sequence $$\left\{2,\frac{2}{3},\frac{4}{15},\frac{4}{35},\frac{16}{315},\frac{16}{693}, \frac{32}{3003},\frac{32}{6435},\frac{256}{109395},\frac{256}{230945},\frac{512}{969969}\right\}$$ and the partial sums the sequence $$\left\{2,\frac{8}{3},\frac{44}{15},\frac{64}{21},\frac{976}{315},\frac{10816} {3465},\frac{141088}{45045},\frac{47104}{15015},\frac{2404096}{765765},\frac{45693952}{14549535}\right\}$$ It does not converge very fast sine $$\frac{a_{n+1}}{a_n}=\frac{n+1}{2 n+3}=\frac{1}{2}-\frac{1}{4 n}+O\left(\frac{1}{n^2}\right)$$

A better one is $$\pi=\sum_{n=0}^\infty \frac{ (5 n+3)}{2^{n-1}\,(3 n+1) (3 n+2) \binom{3 n}{n}}$$ The terms form the sequence $$\left\{3,\frac{2}{15},\frac{13}{1680},\frac{3}{6160},\frac{23}{720720},\frac{ 1}{466752},\frac{11}{75246080},\frac{1}{99095040}\right\}$$ and the partial sums the sequence $$\left\{3,\frac{47}{15},\frac{1759}{560},\frac{2419}{770},\frac{205837}{65520} ,\frac{153966181}{49008960},\frac{23402860601}{7449361920},\frac{989459183 }{314954640}\right\}$$ It converges much faster that the previous one since

$$\frac{a_{n+1}}{a_n}=\frac{(n+1) (2 n+1) (5 n+8)}{3 (3 n+4) (3 n+5) (5 n+3)}= \frac{2}{27}-\frac{1}{27 n}+O\left(\frac{1}{n^2}\right)$$

score 3 · Answer 2 · edited Dec 28 '20 at 01:09

3

We know what the Taylor series for $\tan^{-1}(x)$ is the following.

$$ \tan^{-1}(x)=\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} $$

We also know that $\dfrac{\pi}4 = \tan^{-1}(1)$. So:

$$ \pi = \sum_{n=0}^{\infty} (-1)^n \dfrac{4}{2n+1} $$

So you can just take the sequence of rational numbers that converge to $\pi$ as:

$$ a_N = \sum_{n=0}^{N} (-1)^n \dfrac{4}{2n+1} $$

This follows the same comment mentioned by @JMoravitz.

edited Dec 28 '20 at 01:09

Angelo

12,328

answered Dec 28 '20 at 00:58

Soroush khoubyarian

1,072

2

May be better to explicitly work it out for a non-alternating all positive series: $$ a_N = \sum_{n=0}^{N} \dfrac{8}{(4n+1)(4n+3)} $$ – Deepak Dec 28 '20 at 01:15
Oh correct. They only mentioned the sequence of rationals has to be positive, but I think they actually meant the series should be non-alternating. – Soroush khoubyarian Dec 28 '20 at 01:17

Jaume Oliver Lafont · Answer 3 · 2021-02-10T18:16:57.680

0

We have

$$\pi = 3 + \sum_{k=0}^\infty \frac{24}{(4 k + 2) (4 k + 3) (4 k + 5) (4 k + 6)}$$

$$ = 3 +\frac{2}{15}+\frac{2}{315}+\frac{6}{5005}+\frac{2}{5355}+\frac{2}{13167} + \frac{6}{82225} + ...$$ which is equivalent to the integral

$$ 2 \int_0^1 \frac{x(1-x)^2}{1+x^2} dx = \pi - 3 $$

edited Feb 10 '21 at 18:16

answered Feb 10 '21 at 18:09

Jaume Oliver Lafont

5,212

Is there a positive series of rational numbers that converge to $\pi$?

3 Answers3