You don't need to solve it do you ?
It isn't as if you need to know the value of n to say 3 decimal places.
You're looking for an integer value for n, so a simple crude trial and error substitution will do.
The goal is to show how many terms of the series 1/n*2^n are needed to approximate ln 2 with 6 digit accuracy.
So I set
1/(2^(n+1)*(n+1)<10^-6
And now this means
2^n*(n+1)>500,000
I know the answer is supposed to be 17, but the brain fade is in solving this inequality (unless I'm doing the whole problem wrong). I know it should be easy, but I'm at a little loss here.
Thanks.
6 digit accuracy requires rounding, so half of the above is 0.0000005
invert
ln(2000000)=14.5087
ln(2) = 0.6931
Here is "a simple crude trial and error substitution" method.
n=1 : ln(n)= 0.00 ::
n=20.93 : ln(20.93)=3.04 ::
n=16.55 : ln(16.55)=2.81 ::
n=16.88 : ln(16.88)=2.83 ::
n=16.85 : ln(16.85)=2.824 ::
The required value of n (rounded to whole number) is 17.
:
.