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If $$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$ find $f''(-2)$.


I know that, by ratio test, the previous sum converges iff $x\in(0,6)$.

I did:

$$\begin{matrix} f'(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}\right)'&=& \displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}n{(x-3)}^{n-1}}&=& \displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}} \\ f''(x)&=&\left(\displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}{(x-3)}^{n-1}}\right)'&=& \displaystyle\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{3^n}(n-1){(x-3)}^{n-2}}, \end{matrix}$$

so I did the radio test on the last sum:

$$\begin{matrix} &&\displaystyle\lim_{n\to\infty}{\left|\dfrac{a_{n+1}}{a_n}\right|} \\ &=&\displaystyle\lim_{n\to\infty}{\left|\dfrac{{(-1)}^{n+2}n{(x-3)}^{n-1}}{3^{n+1}}\dfrac{3^n}{{(-1)}^{n+1}(n-1){(x-3)}^{n-2}}\right|}\\ &=&\dfrac{|x-3|}{3}\underbrace{\displaystyle\lim_{n\to\infty}{\left|\dfrac{n}{n-1}\right|}}_{=\:1}\\ &\Rightarrow&|x-3|<3\\ &\Rightarrow&x\in(0,6),& \end{matrix}$$

and since $-2\not\in(0,6)$ we cannot find $f''(-2)$.


First question: is my reasoning right?

If yes, to my amazement the radius of convergence of both functions are the same. Out of curiosity, why is this happening? The convergence can be generalize to $n$-th derivative?

Thank you!

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  • $\begingroup$ @Crostul but WolframAlpha says another thing... $\endgroup$ – manooooh Jul 4 '18 at 22:16
  • $\begingroup$ Crostul's result holds under condition that we are inside the range of convergence. Outside of it, it's a value you can assign to the sum in the generalized fashion (similar to what you can do to the famous 1+2+3+...=-1/12 sum), but technically, it doesn't converge so it has no value. $\endgroup$ – orion Jul 4 '18 at 22:20
  • $\begingroup$ Ah... I love the $1+2+3+\cdots=-1/12$. Ok. $\endgroup$ – manooooh Jul 4 '18 at 22:21
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Yes, it always works. It can be proved (this is a standard result about power series) that if the power series $\displaystyle\sum_{n=0}^\infty a_n(z-a)^n$ has radius of convergence $R$ then;

  1. the radius of convergence of the pwer series $\displaystyle\sum_{n=1}^\infty na_n(z-a)^{n-1}$ is also $R$;
  2. $\displaystyle\left(\sum_{n=0}^\infty a_n(z-a)^n\right)'=\sum_{n=1}^\infty na_n(z-a)^{n-1}$.
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  • $\begingroup$ (I think this is the first time you answer one of my questions (I take a look to your answers and they are incredible)) Wow! amazings properties! Thank you. $\endgroup$ – manooooh Jul 4 '18 at 22:20
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    $\begingroup$ @manooooh I'm glad I could help. $\endgroup$ – José Carlos Santos Jul 4 '18 at 22:47
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In terms of finding $f''(-2)$ with $$f(x)=\sum_{n=1}^\infty{\frac{{(-1)}^{n+1}}{n\cdot3^n}{(x-3)}^n}$$ then consider the following.

Method 1

Since \begin{align} f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} = \ln\left(\frac{x}{3}\right) \end{align} then \begin{align} f'(x) &= \frac{1}{x} \\ f''(x) &= - \frac{1}{x^2} \end{align}

Method 2

\begin{align} f(x) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \left(1 - \frac{x}{3}\right)^{n} \\ f'(x) &= \frac{1}{3} \, \sum_{n=1}^{\infty} \left( \frac{3 - x}{3} \right)^{n} = \frac{1}{3 \, \left(1 - \left( 1 - \frac{x}{3} \right) \right)} = \frac{1}{x} \\ f''(x) &= - \frac{1}{x^2}. \end{align}

The value sought can now be obtained by setting $x = -2$.

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