Hello!

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

Before, I showed that the power series represenation of

$\displaystyle f(x)=\frac{1}{3+2x}$

was

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*x^n$

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

\integrate both, and we get

$\displaystyle F(x) =\int\frac{1}{3+2x}dx=ln(3+2x)$

with power series representation

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{x^{n+1}}{n+1} + c$

where c is a constant of integration,

also, we can find the constant of integration c by letting x=0

$\displaystyle ln (3+2*0) = \sum_{n=0}^{\infty}\frac{(-1)^n2^n}{3^{n+1}}*\frac{0^{n+1}}{n+1} + c$

gives us c= ln 3

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

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

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

but c= ln 3, so...

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

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 $\displaystyle \sum{c_n}$ with rational numbers $\displaystyle c_n$"??

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!