$$f(x)=\sum_{k=0}^\infty \frac{2^{-k}}{k+1}(x-1)^k$$
Find the Taylor series for this function around $x=1$.
I don't understand this, isn't that summation already the Taylor polynomial of $f(x)$? If I follow the procedure of Taylor's theorem and differentiate I get the exact same result again.
Also it's a power series and by Taylor's theorem the coefficients $a_k$ should be equal to $f^{(k)}(1)$, which, non-surprisingly, they are.
Taking upon the semi-agreed question as per the comments, your function is:
$$f(x)=\frac{2\ln\left(\frac12(3-x)\right)}{1-x}$$
As per the Taylor series of the natural logarithm:
$$\ln(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$
and the given Taylor series has radius of convergence $R=2$ by the ratio test.