# Math Help - Integration of a Power Series Representation of a Function

1. ## Integration of a Power Series Representation of a Function

Hello!

I am trying to express ln(6) as a series $\sum{c_n}$ with rational numbers $c_n$

Before, I showed that the power series represenation of
$f(x)=\frac{1}{3+2x}$
was
$\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*x^n$

(http://www.mathhelpforum.com/math-he...tml#post401824)

\integrate both, and we get

$F(x) =\int\frac{1}{3+2x}dx=ln(3+2x)$
with power series representation
$\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{x^{n+1}}{n+1} + c$
where c is a constant of integration,

also, we can find the constant of integration c by letting x=0
$ln (3+2*0) = \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{0^{n+1}}{n+1} + c$
gives us c= ln 3

so, we have now $ln6=ln(3)-ln(3/6)$

if we let ln(3/6)=ln(3+2x), we get x=-5/4 (which is in the radius of convergence R=3/2)

I can't seem to put it all together, however, I still can't seem to answet the question, all I have is

ln(6)= ln(3)-ln(3/6)
ln(6)= ln(3) - (the series evaluated at x=-5/4) + ln(3)
where the 2nd ln(3) is the constant of integration

the series evaluated at x=-5/4 is $\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}$

solving for ln(6) gives us $ln(6) = ln (3) + c + \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}$
but c= ln 3, so...

$ln(6) = 2ln(3) +\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{(-5/4)^{n+1}}{n+1}$

yikes! so basically, my question is
is the above expression a suitable answer to the original question,
"I am trying to express ln(6) as a series $\sum{c_n}$ with rational numbers $c_n$"??

sorry if this is not totally clear, I will try and clarify if this doesn't make sense!

Any help is greatly appriciated!
Sorry for being long-winded here!
Thank you!

2. whoops! cleaned up the latex a bit, should make more sense now, sorry!

3. How about this? A commonly known fact is that for $|x|<1$

$\ln(1-x)=-\int\frac{dx}{1-x}=-\int\sum_{n=0}^{\infty}x^n=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\sum_{n=1}^{\infty}\frac{x^n}{n}$

So $\ln(6)=-\ln\left(\frac{1}{6}\right)=-\ln\left(1-\frac{5}{6}\right)=\sum_{n=1}^{\infty}\frac{5^n}{n \cdot 6^n}$