Your series expression for f(x) is right, up to a point. Don't forget that when you develop a power series by integration that you pick up, yes, an arbitrary constant. So that needs to be evaluated. Pick a value for x that will make this easy to deal with, like x = 0. The entire infinite polynomial becomes zero and you just have f(0) = C , so C = ln 2 .
The way you wrote your series otherwise is all right, though many sources would deal with it differently. Rather than removing a factor of x, many people would instead bring the (1/2) under the summation sign, making the general term
(-1)^n \cdot 2^{-(n+1)} \cdot \frac{1}{n+1} \cdot x^{n+1}
Since the index on the infinite sum now starts at 1 , they would next perform an "index shift" by re-assigning (n+1) to n . The index now starts at zero again and the Maclaurin series for our function becomes
ln(x+2) = ln 2 + \Sigma^{\infty}_{n = 0} (-1)^{n-1} \cdot \frac{x^n}{n 2^{n} }
I use a Ratio Test for the radius of convergence, rather than what you used, but I generally agree with your result (though you should write it simply as R = 2 ; radii are non-negative, so you wouldn't write -2 < R < 2 -- I believe you're thinking of the interval of convergence, for which we would still need to test the endpoints...).
The infinite alternating series
1 + (-1) + 1 + (-1) + ...
does not converge by reason of indeterminancy. Since we have a literally infinite reserve of positive and negative 1's, it is possible to use this alternating series to create any integer you like. Thus
0 = 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]
3 = 1 + 1 + 1 + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]
-2 = (-1) + (-1) + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms] , etc.
We can pull out any finite number of positive or negative 1's to produce any desired integer and still be guaranteed total cancellation of "the rest" of the infinity of terms. Such things are in the nature of infinite sets...
