Why the increase in starting value of $n$ when taking derivative of power series?

Question

Why is the starting value of $n$ increased by $1$ when taking derivative of power series?

i.e.:

$f(x) = \sum_\color{red}{n=0}^{\infty} c_{n} x^{n}$

$f'(x) = \sum_\color{red}{n=1}^{\infty} nc_{n} x^{n-1}$

Answer 1

The first term is $c_0x^0 = c_0$, so is constant, so its derivative is zero. If you prefer, you could re-index: $$f'(x) = \sum\limits_{n=0}^\infty(n+1)c_{n+1}x^n,$$ but that's slightly more writing, so most people don't.

Answer 2

Let $$f(x)=a_0+a_1x+a_2x^2+a_3x^3+...=\sum_{n=0}^\infty a_nx^n.$$ Then $$\begin{align} f'(x)&=\frac{d}{dx}\left(a_0+a_1x+a_2x^2+a_3x^3+...\right)\\ &=\frac{d}{dx}a_0+\frac{d}{dx}a_1x+\frac{d}{dx}a_2x^2+\frac{d}{dx}a_3x^3+...\\ &=a_0\frac{d}{dx}1+a_1\frac{d}{dx}x+a_2\frac{d}{dx}x^2+a_3\frac{d}{dx}x^3+...\\ &=a_0\cdot0+a_1\cdot1+a_2\cdot2x+a_3\cdot3x^2+...\\ &=0+1a_1+2a_2x+3a_3x^2+...\\ &=a_1x^{0}+2a_2x^{1}+3a_3x^2+...\\ &=\sum_{n=1}^{\infty}na_nx^{n-1}. \end{align}$$