TITLE: Differentiation of $\sum_{n=0}^{\infty}\frac{1}{(n+1)3^n}x^{n+1}$
QUESTION [9 upvotes]: 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 $2$?

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