Sequnce of rationals less than pi, which converges to pi

[Amer]

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

$$\cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi)$$
for the $$\sqrt{2}$$ i was thinking about the taylor series for $$\sqrt{x}$$
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

 

[I like Serena]

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

$$\cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi)$$
for the $$\sqrt{2}$$ i was thinking about the taylor series for $$\sqrt{x}$$
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]

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

$$\cup (a_{\alpha} , b_{\alpha} ) = (\sqrt{2} , \pi)$$
for the $$\sqrt{2}$$ i was thinking about the taylor series for $$\sqrt{x}$$
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.

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[Amer]

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

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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]

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]

There is always the sequence $\{3,\,3.1,\,3.14,\,3.141,\,3.1415,\,3.14159,\, \ldots\}$.

 

[Nono713]

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 ;)

 

[chisigma]

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 Serena]

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]

you gave me a punch of ideas thanks very much all of you

 

[zzephod]

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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