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sequnce of rationals less than pi, which converges to pi

Amer

Active member
Mar 1, 2012
275
is there any sequnce which converges to [tex]\pi[/tex] such that each term of it less than pi
I know the sequnce related with the taylor expansion of the arctan [tex]\sum_{i=1}^\infty \frac{(-1)^{i+1}}{(2i-1)} [/tex]
but this sequnce first term is bigger than pi
why I am looking for such a sequnce because I want to find [tex]a_{\alpha} , b_{\alpha} [/tex] such that

[tex] \cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi) [/tex]
for the [tex]\sqrt{2} [/tex] i was thinking about the taylor series for [tex]\sqrt{x} [/tex]
but what I am stuck at is the taylor series for a function T(x) is convereges to f(a) if the expansion was around the a

Any ideas
Thanks
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
is there any sequnce which converges to [tex]\pi[/tex] such that each term of it less than pi
I know the sequnce related with the taylor expansion of the arctan [tex]\sum_{i=1}^\infty \frac{(-1)^{i+1}}{(2i-1)} [/tex]
but this sequnce first term is bigger than pi
why I am looking for such a sequnce because I want to find [tex]a_{\alpha} , b_{\alpha} [/tex] such that

[tex] \cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi) [/tex]
for the [tex]\sqrt{2} [/tex] i was thinking about the taylor series for [tex]\sqrt{x} [/tex]
but what I am stuck at is the taylor series for a function T(x) is convereges to f(a) if the expansion was around the a

Any ideas
Thanks
Hi Amer! :)

Well... another sequence is $\zeta(2) = 1 + \dfrac 1 {2^2} + \dfrac 1 {3^2} + ... = \dfrac {\pi^2}{6}$
You can find more sequences like that on wiki.
 

zzephod

Well-known member
Feb 3, 2013
134
is there any sequnce which converges to [tex]\pi[/tex] such that each term of it less than pi
I know the sequnce related with the taylor expansion of the arctan [tex]\sum_{i=1}^\infty \frac{(-1)^{i+1}}{(2i-1)} [/tex]
but this sequnce first term is bigger than pi
why I am looking for such a sequnce because I want to find [tex]a_{\alpha} , b_{\alpha} [/tex] such that

[tex] \cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi) [/tex]
for the [tex]\sqrt{2} [/tex] i was thinking about the taylor series for [tex]\sqrt{x} [/tex]
but what I am stuck at is the taylor series for a function T(x) is convereges to f(a) if the expansion was around the a

Any ideas
Thanks
Archimedes sequence of inscribed regular polygons to a circle of unit diameter will give what you want.

.
 

Amer

Active member
Mar 1, 2012
275
Archimedes sequence of inscribed regular polygons to a circle of unit diameter will give what you want.

.
can you give me a link about what you are talking about, thanks again
I googled Archimedes sequnce but i did not get trivial thing
 

Amer

Active member
Mar 1, 2012
275
Hi Amer! :)

Well... another sequence is $\zeta(2) = 1 + \dfrac 1 {2^2} + \dfrac 1 {3^2} + ... = \dfrac {\pi^2}{6}$
You can find more sequences like that on wiki.
interesting link, thanks
I am looking for pi without power :)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
There is always the sequence $\{3,\,3.1,\,3.14,\,3.141,\,3.1415,\,3.14159,\, \ldots\}$.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Or more generally, any sequence $a_i$ which satisfies:

$$a_i = \left \lfloor \left ( \pi - f(i) \right ) \cdot b^i \right \rfloor b^{-i}$$

For any function $f$, with $f(i) \geq 0$, such that the limit to infinity of $f$ be zero, and any rational base $b > 1$.

The sequence Opalg presented is one of the family above, with $b = 10$ and $f(i) = 0$. For instance, in base $4$:

$$\{ 3,\ 3,\ 3.125,\ 3.140625,\ \dots \}$$

Using a non-constant function, say, $f(i) = \pi e^{-i}$, we get, in base 10:

$$\{ 0,\ 1.9,\ 2.71,\ 2.985,\ 3.084,\ 3.12042,\ \dots \}$$

But of course, using this "digit extraction" method is probably not Amer wanted ;)
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Any alternating sign terms series can be transformed in a positive terms series as follows...


$\displaystyle \sum_{n=0}^{\infty} (-1)^{n}\ a_{n} = (a_{0}-a_{1}) + (a_{2}-a_{1}) +...$ (1)


... so that is...


$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}} {2n + 1} = (1-\frac{1}{3}) + (\frac{1}{5}- \frac{1}{7}) +...$ (2)

Kind regards


$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Any alternating sign terms series can be transformed in a positive terms series as follows...


$\displaystyle \sum_{n=0}^{\infty} (-1)^{n}\ a_{n} = (a_{0}-a_{1}) + (a_{2}-a_{1}) +...$ (1)


... so that is...


$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}} {2n + 1} = (1-\frac{1}{3}) + (\frac{1}{5}- \frac{1}{7}) +...$ (2)

Kind regards


$\chi$ $\sigma$
I like it! (Sun)

Cleaning it up a little so we don't see that "ugly" minus sign anymore, you'd get:
$8 \displaystyle\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+3)} = 8(\frac{1}{1 \cdot 3} + \frac{1}{5 \cdot 7} + \frac{1}{9 \cdot 11} + ...) = 8(\frac{1}{3} + \frac{1}{35} + \frac{1}{99} + ...)= \pi$​

Looks just as if there never was a minus sign involved! (Wink)
 

Amer

Active member
Mar 1, 2012
275
you gave me a punch of ideas thanks very much all of you
 

zzephod

Well-known member
Feb 3, 2013
134
can you give me a link about what you are talking about, thanks again
I googled Archimedes sequnce but i did not get trivial thing
A paper (pdf) that provides the background and the formula using only elementary methods can be found >>here<<

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