Problem with Power Series Problem So, I was given this problem:
Let f(x) be the function defined by the power series:
$$1-\frac{1}{3}(x-2)+\frac{1}{9}(x-2)^2-\frac{1}{27}(x-2)^3+...+(-1)^n\frac{(x-2)^n}{3^n}$$
(a) For what values of x does the series converge?
(b)Write the first four terms and the general term of the power series for f ’(x)
centered at x = 2 and find f ’(2).
(c) Write an equation of the tangent line of f(x) at the point where x = 2.
So far, I solved for a:
x - 2 < 3 --> -1 < x < 5
However, I'm struggling with B. So far I've done:
$$f'=0-\frac{1}{3}+\frac{1}{9}(2x-4)-\frac{(x-2)^2}{9}+\frac{4}{81}(x-2)^3$$
I'm not sure how to find the general term nor do I know where to start with C. Any help?
 A: I suppose that you're trying to says
$$1-\frac{1}{3}(x-2)+\frac{1}{9}(x-2)^{2}-\frac{1}{27}(x-2)^{3}+\cdots+(-1)^{n}\frac{(x-2)^{n}}{3^{n}}+\cdots$$
Otherwise, if you don't add the points "$+\cdots$"  after the $n$-th term, what you indicate is just a polynomial of degree less than or equal to $n$, but you're saying about a power series.
Well, notice what we can re-write your series as
$$\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{3^{n}}(x-2)^{n}$$
This is a power series with $n-$th term $a_{n}=\frac{(-1)^{n}}{3^{n}}$ and centered at $x=2$.

*

*By ratio test,
\begin{align*}
\rho&=\lim_{n\to +\infty}\left|\frac{\frac{(-1)^{n+1}}{3^{n+1}}(x-2)^{n+1} }{\frac{(-1)^{n}}{3^{n}}(x-2)^{n}}\right|,\\&=\lim_{n\to +\infty}\frac{|x-2|}{3},\\&=\frac{|x-2|}{3}
\end{align*}
Now, consider the cases a) $\rho<1$, b)$\rho>1$ and c) $\rho=1$ for the test the convergence.


Just a small remark: you need to check all the cases: a), b) and c) in the way to study the interval of convergence for $x$.


*

*Over the interval of convergence we have,
$$f(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{3^{n}}(x-2)^{n}$$Hence,$$f'(x)=\frac{{\rm d}f}{{\rm d}x}=\sum_{\color{red}{n=1}}^{+\infty}\frac{(-1)^{n}}{3^{n}}\color{blue}{n(x-2)^{n-1}}.$$

Just a small remark: Notice that when calculating the first derivative of $f$, we move the index $1$ unit (in red and yes, if asked for the second derivative, move the index forward by $2$ units, etc.) and also note that the derivative is calculated with respect to $x$ (the color blue) as usual for a derivative of $x^n$.

Then setting $x=2$ you can obtain $f'(2)$.

*

*Consider the equation
$$y-y_{0}=f'(x_{0})(x-x_{0})$$
with $x_{0}=2, y_{0}=f(x_{0})=f(2)$ and $f'(x_{0})=f'(0)$ and then find the tangent-line.

