I have the following exam Question:
Consider the power series
$$P_\infty(x) = \sum_{k=1}^{\infty} \dfrac{-(x-1)^k}{2k}$$
- Examine $P_\infty(x)$ for convergence.
- Determine the Taylor polynomial $T_{2}(x)$ of the function $f : \mathbb{R}_{\lt2} \to \mathbb{R}$ with $f(x) = \ln{\sqrt{2 - x}}$ and $x_0 = 1$
- Show that $P_\infty(x)$ is the Taylor series of function $f$ at point $x_0 = 1$
Some verification for steps 1 and 2 would be nice, however as the title indicates, my main query concerns part 3.
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Progress so far:
For part 1, it seems that this could be achieved via application of the ratio test. Here I am seeking verification that my application of this is correct, and the limit can be calculated as follows:
\begin{align*} L &= \lim_{k\to\infty} \Big| \dfrac{\frac{-(x-1)^{k+1}}{2(k+1)}}{\frac{-(x-1)^{k}}{2k}} \Big| \\ \vdots \\ &= |1 - x| \\ ~ \\ \end{align*}
Thus, $f$ can be said to be convergent if $|x - 1| < 1$, $\hspace{0.4 cm}$ or $\hspace{0.4 cm} 0 < x < 2$.
For part 2, I based my calculations off the formula provided by this link:
$$T_2(x) = f(x_0) + f'(x_0)(x - x_0) + \dfrac{f''(x_0)}{2!} \cdot (x-x_0)^2$$
After checking my answer on Wolfram, I concluded that $$T_2(x) = \dfrac{(1-x)}{2} - \dfrac{(x-1)^2}{4} $$
For part 3, I have tried to copy the method which was provided in this related question (namely, calculate furter terms of the Taylor polynomial until they can be generalized). After calculating the next term of the polynomial to be $f^3(x_0) = -\frac{1}{2}$, I concluded that $f^k(x_0) = -\frac{1}{2}$. Give this, we should have all the ingredients from the Taylor Polynomial to generalize these as the given power series, that is to say the factorial that occurs will be represented in our sum as
$\frac{1}{k!} = \sum\limits_{k=1}^{\infty} \frac{1}{k}$
and the part $(x-1)^k$ is already contained.
I just don't know how to present this in a mathematical way. Can someone please provide some assistance in formalizing my solution?
For part $3$ you may consider the following approach.
$$\ln \sqrt {2-x} =(1/2) \ln (2-x) = -1/2 \int _1 ^x \frac {dt}{2-t} $$
$$ \frac {1}{2-t} = \frac {1}{1-(t-1)} = 1+(t-1)+(t-1)^2+(t-1)^3+...$$
Upon integration you get the desired infinite power series about $x_0=1$
Now you apply the uniqueness of Taylor Series to get the result.