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Let $$f(x)=\sum_{n=0}^{\infty}\frac{1}{(n+1)3^n}x^{n+1}=x+\frac{x^2}{6}+\frac{x^3}{27}+\frac{x^4}{108}+\cdots$$

The question asked me to use the knowledge of series to compute $f'(2)$.

How should I solve? Wouldn't be just differentiate each term and substitute $2$?

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Yes, that can be one way. You will get a geometric progression:

$$f'(x) = 1+ \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} ...$$

So $f'(x) = \frac{1}{1-x/3}$, or $f'(2) = 3$.

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$$\begin{align} f(x)&=\sum_{n=0}^\infty\frac 1{(n+1)3^n}x^{n+1}\\ f'(x)&=\frac d{dx}\sum_{n=0}^\infty\frac 1{(n+1)3^n}x^{n+1}\\ &=\sum_{n=0}^\infty\frac d{dx}\frac 1{(n+1)3^n}x^{n+1} &&\text{(by Fubini/Tonelli's theorem)}\\ &=\sum_{n=0}^\infty\frac 1{(n+1)3^n}\cdot (n+1)x^n\\ &=\sum_{n=0}^\infty \left(\frac x3\right)^n\\ &=\frac 1{1-\frac x3}\\ \therefore f'(2)&=3 \end{align}$$

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    $\begingroup$ How do you justify interchanging the differentiation with the summation? There's at least something to be said there, even if it's only quoting the appropriate theorem about radii of convergence. $\endgroup$ – Mark Dickinson Dec 11 '17 at 21:35
  • $\begingroup$ @MarkDickinson - Theorem quoted! $\endgroup$ – hypergeometric Dec 12 '17 at 16:40
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$$-f(x)/3=\ln(1-x/3)$$ for $-1\le x/3<1$

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