# Differentiation of $\sum_{n=0}^{\infty}\frac{1}{(n+1)3^n}x^{n+1}$

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 $2$?

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 $f'(2) = 3$.

\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}

• 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. Dec 11 '17 at 21:35
• @MarkDickinson - Theorem quoted! Dec 12 '17 at 16:40

$$-f(x)/3=\ln(1-x/3)$$ for $-1\le x/3<1$