I am trying to express ln(6) as a series with rational numbers
Before, I showed that the power series represenation of
\integrate both, and we get
with power series representation
where c is a constant of integration,
also, we can find the constant of integration c by letting x=0
gives us c= ln 3
so, we have now
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) - (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
solving for ln(6) gives us
but c= ln 3, so...
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 with rational numbers "??
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!