# √2 as a sum of infinite series

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• Feb 5th 2011, 11:42 AM
ThodorisK
√2 as a sum of infinite series
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.
• Feb 5th 2011, 11:50 AM
Plato
Quote:

Originally Posted by ThodorisK
I cant' find any. And some terms of the continued fraction e.g. the fourth term 1+(5/12), are greater than √2.

Do you simply a series the sum of which is $\displaystyle \sqrt2~?$
• Feb 5th 2011, 11:52 AM
TheCoffeeMachine
$\displaystyle \displaystyle \sum_{k=0}^\infty \frac{(-1)^k(2k)!}{(1-2k)(k!)^2(4^k)}.$
• Feb 5th 2011, 11:58 AM
earboth
Quote:

Originally Posted by ThodorisK
I cant' find any.

And some terms of the continued fraction e.g. the fourth term 1+(5/12), are greater than √2.

1. Use the function $\displaystyle f(x)=\sqrt{x}$

2. Develop f in a Taylor serie:

$\displaystyle f(a+h)=f(a)+\frac h{1!} f'(a) + \frac{h^2}{2!} f''(a) + ... + \frac{h^{n-1}}{(n-1)!} f^{(n-1)}(a) + R_n$

wit a = 1 and h = 1
• Feb 5th 2011, 12:53 PM
chisigma
$\displaystyle \displaystyle \sqrt{2} = 2\ \cos \frac{\pi}{4}} = 2\ \sum_{n=0}^{\infty} (-1)^{n} \frac{(\frac{\pi}{4})^{2n}}{(2n)!}$
• Feb 5th 2011, 12:58 PM
mr fantastic
I suspect the OP wants to express $\displaystyle \sqrt{2}$ as a continued fraction. But until clarification of the question is provided, there is no point stabbing in the dark at possbile answers.
• Feb 5th 2011, 05:41 PM
Soroban
Hello, ThodorisK!

If you want a continued fraction for $\displaystyle \sqrt{2}$, here's one . . .

$\displaystyle \displaystyle \sqrt{2} \;=\;1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \hdots}}}}}$

• Feb 5th 2011, 11:45 PM
ThodorisK
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.
• Feb 6th 2011, 12:53 AM
tonio
Quote:

Originally Posted by ThodorisK
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.

Also, I want it to contain numbers only. Or directions of how the numbers are combined, e.g. not something with π in it.

Your last words seem to imply that you're not really sure either of what you

want or else you're not quite sure what you were asked to do: the "something with n in it" is numbers in

the examples you've been given, and the n only serves as counting index.

A simple way to achieve what you want is:

$\displaystyle \displaystyle{1=\sum\limits^\infty_{n=1}\left(\fra c{1}{2}\right)^n\Longrightarrow \sqrt{2}=\sum\limits^\infty_{n=1}\frac{1}{2^{n-1/2}}=\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{1 }{4\sqrt{2}}+\ldots}$

Tonio
• Feb 6th 2011, 02:58 AM
Do you mean an infinite sequence of sums ?

$\displaystyle \displaystyle\ a_{n+1}=\frac{a_n}{2}+\frac{1}{a_n},\;\;\;\;a_0=1$

gives the sequence

$\displaystyle \displaystyle\ 1,\;\;\;\frac{1}{2}+1=\frac{3}{2},\;\;\;\frac{3}{4 }+\frac{2}{3}=\frac{17}{12}=1+\frac{5}{12},....... ........$

though you say that should be the 4th term...
• Feb 6th 2011, 03:34 AM
mr fantastic
Quote:

Originally Posted by ThodorisK
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.

We are still in the dark.

State the question exactly as it is written in your book (or wherever you got it from).
• Feb 6th 2011, 04:02 AM
ThodorisK
Let's say that I want something like the 1+2(1/8)+4(1/8)^2+8(1/8)^2+...of the area of the parabola
Geometric series - Wikipedia, the free encyclopedia
where instead of the area of the parabola, it is the side √2 of a triangle with sides 1,1,√2. So, none of the terms of the sum can be greater than √2. What I mean by "the terms of the sum" is for example regarding the continued fraction:

first term = 1
second term = 1+1/2
third term = 1+(1/(2+(1/2))=1+(2/5)
• Feb 6th 2011, 10:00 AM
wonderboy1953
Binomial theorem
Quote:

Originally Posted by ThodorisK
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.

I seem to recall that you can express $\displaystyle \sqrt{2}$ as a series using the binomial theorem. What I can do for you is when I get home, I can look it up and get back to you tomorrow on it.
• Feb 6th 2011, 10:07 AM
skeeter
Square_root_of_2

scroll down to Series and product representations
• Feb 6th 2011, 11:49 PM
CaptainBlack
Quote:

Originally Posted by ThodorisK
I dont want THE continued fraction because it's not formed as a sum of infinite series. Also, when I try to make it as such, its 2nd, 4th, 6th and so on terms are greater than √2, e.g. the fourth term is 1+(5/12), so I guess it's not what I want, but I might be wrong.

Do not modify your original post in response to something in a reply, but post a new message quoting what you are responding to and your response. That way the thread will make sense and can be followed by new viewers.

CB
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