Infinite series (7)

**NonCommAlg** · December 7th 2009, 08:25 PM

Define the sequence $\{x_n \}$ by $x_1=3$ and $x_{n+1}=x_n^2 - 2, \ \ n \geq 1.$ Evaluate $\sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}.$

**simplependulum** · December 8th 2009, 01:56 AM

wow !

How can people ( is that you ?) find out this this expression/series for irrational numbers ?

we have

$x_1 = 3 , x_2 = 7 , x_3 = 47 , x_4 = 2207 , x_5 = 4870847 ...$

we can see that the series converges so quickly !

for this problem ,

I use a quite complicated method , is there any method better than mine ?

Go through it !

the general term of the seqence is $2 \cosh(2^{n-1}y)$

where $y = \cosh^{-1}(x_1 /2 )$

so the product of first $n$ terms is

$x_1 x_2 x_3 .. x_n =$

$2^n \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^n}{\sinh(y)} \cdot \sinh(y) \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^{n-1}}{\sinh(y)} \cdot \sinh(2y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^{n-2}}{\sinh(y)} \cdot \sinh(4y) \cosh(4y) ... \cosh(2^{n-1}y) = \frac{\sinh(2^n y)}{ \sinh(y) }$

the series can be written in this form :

$2\sinh(y) \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

$\sqrt{x_1^2 - 4 } \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

To evluate this series , we just need to create $\frac{1}{a^2 - 1} - \frac{1}{a^2 - 1}$ next to the series .

I got

$\sqrt{ x_1^2 - 4 } \left( \frac{1}{ e^{2y} - 1} - \frac{1}{ e^{ 2^{m+1} y } - 1 } \right )$

$e^y = \frac{ x_1 + \sqrt{ x_1^2 - 4 }}{2}$

for $x_1 = 3 , m \to \infty$

I got $\frac{ 3 - \sqrt{5} }{2}$

or an elegant-appearance

$\sqrt{5} = 3 - \frac{2}{3} - \frac{2}{3\cdot7} - \frac{2}{3 \cdot 7 \cdot 47 } - \frac{2}{3\cdot7\cdot47\cdot2207} - ..$

the fifth term is about $10^{-6}$ , wow

**NonCommAlg** · December 8th 2009, 02:44 AM

Originally Posted by simplependulum

wow !

How can people ( is that you ?) find out this this expression/series for irrational numbers ?

we have

$x_1 = 3 , x_2 = 7 , x_3 = 47 , x_4 = 2207 , x_5 = 4870847 ...$

we can see that the series converges so quickly !

for this problem ,

I use a quite complicated method , is there any method better than mine ?

Go through it !

the general term of the seqence is $2 \cosh(2^{n-1}y)$

where $y = \cosh^{-1}(x_1 /2 )$

so the product of first $n$ terms is

$x_1 x_2 x_3 .. x_n =$

$2^n \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^n}{\sinh(y)} \cdot \sinh(y) \cosh(y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^{n-1}}{\sinh(y)} \cdot \sinh(2y) \cosh(2y) \cosh(4y) ... \cosh(2^{n-1}y)$

$= \frac{2^{n-2}}{\sinh(y)} \cdot \sinh(4y) \cosh(4y) ... \cosh(2^{n-1}y) = \frac{\sinh(2^n y)}{ \sinh(y) }$

the series can be written in this form :

$2\sinh(y) \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

$\sqrt{x_1^2 - 4 } \sum_{k=1}^{m} \frac{ a^{2^n} }{ a^{2^{n+1}} - 1 }$

To evluate this series , we just need to create $\frac{1}{a^2 - 1} - \frac{1}{a^2 - 1}$ next to the series .

I got

$\sqrt{ x_1^2 - 4 } \left( \frac{1}{ e^{2y} - 1} - \frac{1}{ e^{ 2^{m+1} y } - 1 } \right )$

$e^y = \frac{ x_1 + \sqrt{ x_1^2 - 4 }}{2}$

for $x_1 = 3 , m \to \infty$

I got $\frac{ 3 - \sqrt{5} }{2}$

or an elegant-appearance

$\sqrt{5} = 3 - \frac{2}{3} - \frac{2}{3\cdot7} - \frac{2}{3 \cdot 7 \cdot 47 } - \frac{2}{3\cdot7\cdot47\cdot2207} - ..$

the fifth term is about $10^{-6}$ , wow

your solution looks a little too complicated but it seems ok and your final answer is correct. well-done! i'll post my solution later. let's see if anybody else would like to try it.

**NonCommAlg** · December 12th 2009, 03:28 PM

here's another approach: the series $\sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}$ is a "telescoping" series because $\frac{1}{x_1x_2 \cdots x_n} = \frac{1}{2} \left( \frac{x_n}{x_1 x_2 \cdots x_{n-1}} - \frac{x_{n+1}}{x_1x_2 \cdots x_n} \right).$

**simplependulum** · December 12th 2009, 07:25 PM

Originally Posted by NonCommAlg

here's another approach: the series $\sum_{n=1}^{\infty} \frac{1}{x_1x_2 \cdots x_n}$ is a "telescoping" series because $\frac{1}{x_1x_2 \cdots x_n} = \frac{1}{2} \left( \frac{x_n}{x_1 x_2 \cdots x_{n-1}} - \frac{x_{n+1}}{x_1x_2 \cdots x_n} \right).$

I wanna know how you evaluate the limit :

$\lim_{n\to\infty} \frac{ x_{n+1}}{ x_1 x_2 x_3 ... x_n }$

my method is to multiply $\sqrt{ x_1^2 - 4 }$ both the numerator and the denominator ...

**simplependulum** · December 12th 2009, 07:30 PM

I have consider another method but got stuck somewhere ...

$f(x_1) = \frac{1}{x_1} + \frac{1}{x_1 x_2} + \frac{1}{x_1 x_2 x_3} + ....$

we have

$x_1 f(x_1) = 1 + f(x_2) = 1 + f( x_1^2 - 2 )$

so i obtain a functional equation

$x f(x) = 1 + f(x^2 - 2 )$

and i know $\frac{x}{2}$ is a one of the solution

so i let $f(x) = \frac{x}{2} + g(x)$

then $x g(x) = g(x^2 - 2)$

i found that $g(x) = \sqrt{x^2 - 4 }$

but the problem is here :

$k \sqrt{ x^2 - 4 }$ is also the solution but how do we get the exact value of $k$ ? ( $k = -\frac{1}{2}$ )

**NonCommAlg** · December 12th 2009, 09:44 PM

here's the trick: $x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4$ and thus $x_{n+1}^2 - 4 = x_n^2(x_n^2-4).$ therefore $x_{n+1}^2-4=x_n^2x_{n-1}^2(x_{n-1}^2 - 4)=x_n^2 x_{n-1}^2x_{n-2}^2(x_{n-2}^2 - 4) = \cdots$ got it?

**simplependulum** · December 12th 2009, 10:29 PM

Originally Posted by NonCommAlg

here's the trick: $x_{n+1}^2=(x_n^2 - 2)^2=x_n^2(x_n^2 - 4) + 4$ and thus $x_{n+1}^2 - 4 = x_n^2(x_n^2-4).$ therefore $x_{n+1}^2-4=x_n^2x_{n-1}^2(x_{n-1}^2 - 4)=x_n^2 x_{n-1}^2x_{n-2}^2(x_{n-2}^2 - 4) = \cdots$ got it?

oh , i see .

this method is similar to my method , $\sqrt{5}$ is the limit .

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