calculating the Taylor series remainder of $\ln(2)$ I'm trying to calculate the remainder for the Taylor series expansion of $\ln(2)$, and would greatly appreciate your support.
I've learnt that when expanding the Taylor series:
$f(x) = f(a) + \frac{f'(a)}{1!}(x-a)... + \frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_n(x)$
Where $R_n(x)$ is the remainder, s.t. $R_n(x)=\frac{f^{(n)}(a+\theta(x-a))}{n!}(x-a)^n$, given $0 <\theta<1$
My assumption was that using  the series $\ln(1+x)$ and replacing it by $2$ would guide me to a solution.
$\ln(1+x) = x-\frac{1}{2}x^2+\frac{1}{3}x^3...$
Though, that's not the right expansion. How would I correctly expand this, thanks for the help.
 A: Hint:
As $2=\frac1{1/2}$, you can use that
$$\ln 2=-\ln\Bigl(1-\frac12\Bigr),$$
and use the expansion of $\ln(1-x)$.
A: You actually can use the expansion $\log (1+x) = x- \frac {x^2}{2}+\frac {x^3}{3}...$, which is the correct one for $\log (1+x) \; \;$ when $\; \; -1 \lt x \le 1$.
To estimate the remainder using the integral form might be easier:
$$\int _0^x \frac {t^n}{1+t} dt \le \int _0^x t^n dt = \frac {x^{n+1}}{n+1} $$
So when $x=1$ (there the expantion holds but not further), we can make the remainder as small as we want it to be by choosing a suitable large $n$.

EDIT: I gave you the solution, but I forgot to explain the process of getting there, silly me...
Remember that $\log (1 + x) = \int_0^x \frac {1}{1+t}$ and also consider the equation:
$$\frac {1}{1+t} =\frac {1 + t - t}{1+t} = 1 + \frac {-t}{1+t} = 1- \frac {t^2 +t - t^2}{t+1} = 1 - \frac {t(t+1) - t^2}{t+1} = 1 - t + \frac {t^2}{t+1}= 1- t + \frac {t^3 + t^2 - t^3}{t+1} = 1 - t + \frac {t^2(t+1) - t^3}{t+1} = 1 - t + t^2 - \frac {t^3}{t+1} = ...$$
A: $$\sum_{k=0}^{n-1}(-x)^k=\frac1{1+x}-\frac{(-x)^n}{1+x}$$
and after integration the exact remainder is
$$\int_0^x\frac{t^n}{1+t}dt$$ (sign omitted). This is an incomplete Beta integral.
