# series problem

• Aug 18th 2006, 07:57 AM
parallel
series problem
hello

I need to find the sum:

sigma 0-->infinity (-1/3)^n * 1/(2n+1)

i looked at the power series : x^2n (1)
the sum of this series is : 1/(1-x^2) (2)

then by integration of (1) I got : x^2n+1 * 1/(2n+1) (3)
by integration of (2) i got the sum of (3)

I dont know what to do now

help would be appreciated.
• Aug 18th 2006, 09:20 AM
tize
I'm not sur but a primitive of 1/(1-x^2) is Argth(x) so by integration we have :
\sum_{n=0}to{infty} (x^(2n+1))/(2n+1) = Argth(x) so
x.\sum_{n=0}to{infty} (x^(2n))/(2n+1) = Argth(x)
x.\sum_{n=0}to{infty} ((x^2)^n)/(2n+1) = Argth(x). Take x= i/sqrt(3) such that x^2= -1/3 and you have :
i/sqrt(3).\sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3)) and finaly :
\sum_{n=0}to{infty} ((-1/3)^n)/(2n+1) = Argth(i/sqrt(3))/(i/sqrt(3))

I hope I didn't make mistake.
• Aug 18th 2006, 09:26 AM
CaptainBlack
Quote:

Originally Posted by parallel
hello

I need to find the sum:

sigma 0-->infinity (-1/3)^n * 1/(2n+1)

i looked at the power series : x^2n (1)
the sum of this series is : 1/(1-x^2) (2)

then by integration of (1) I got : x^2n+1 * 1/(2n+1) (3)
by integration of (2) i got the sum of (3)

I dont know what to do now

help would be appreciated.

Consider the series expansion:

$
\arctan(x)=\sum_0^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1},\ \ |x|<1
$
,

so:

$
\arctan(\sqrt{x})=\sum_0^{\infty}\frac{(-1)^n x^{(2n+1)/2}}{2n+1}$

$
=x^{1/2}\sum_0^{\infty}\frac{(-x)^{n}}{2n+1},\ \ |x|<1
$
and the rest should follow easily.

RonL
• Aug 18th 2006, 10:52 AM
parallel
tize:

the primitive of 1/1+x^2 is arctan(x) :) the primitive of 1/(1-x^2) is:
1/2 * ln[ (1+x) / (1-x) ]
but I understood the idea of what you were trying to do.

but still,i can't get the result as it should be

CaptainBlack

do you mean that the sum is just arctan(1/sqrt(3))?

i'm using an online calculator, the result should be 0.90~.
and as I said before,I can't get the result,can you please help me with my solution too?

i got that the sum is: i/sqrt(3) * 1/2 * ln[ (1+x) / (1-x) ] (x=i/sqrt(3))
• Aug 18th 2006, 01:13 PM
CaptainBlack
Quote:

Originally Posted by parallel
tize:

the primitive of 1/1+x^2 is arctan(x) :) the primitive of 1/(1-x^2) is:
1/2 * ln[ (1+x) / (1-x) ]
but I understood the idea of what you were trying to do.

but still,i can't get the result as it should be

CaptainBlack

do you mean that the sum is just arctan(1/sqrt(3))?

i'm using an online calculator, the result should be 0.90~.
and as I said before,I can't get the result,can you please help me with my solution too?

i got that the sum is: i/sqrt(3) * 1/2 * ln[ (1+x) / (1-x) ] (x=i/sqrt(3))

First there is an error in my handbook (Scham's) so rather than what I gave
the relevant expansion for $\arctan$ is:

$
\arctan(x)=\sum_0^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1},\ \ |x|<1
$
,

so

$
\arctan(\sqrt{x})=\sum_0^{\infty}(-1)^n\frac{x^{(2n+1)/2}}{2n+1}
$

$
=x^{1/2}\sum_0^{\infty}\frac{(-x)^{n}}{2n+1},\ \ |x|<1
$

So:

$
\sum_0^{\infty}\frac{(-x)^{n}}{2n+1}=\frac{\arctan(\sqrt{x})}{\sqrt{x}},
\ \ |x|<1
$
.

Hence:

$
\sum_0^{\infty}\frac{(-1/3)^{n}}{2n+1}=\frac{\arctan(\sqrt{1/3})}{\sqrt{1/3}}\approx 0.9069
$
.

RonL
• Aug 19th 2006, 02:26 AM
parallel