Derivative of Sereis

Question

I was looking at the series for $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ and for $\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$. I noticed that the second sum is the derivative of the first Since $$\frac{d}{dx}x^n=nx^{n-1}$$. Does this mean that $-\frac{2}{(1-x)^3}=\sum_{n=0}^\infty (n-1)nx^{n-2}$ Can Anyone tell me if I am wrong or right?

Answer 1

Note 1: Each time you take a derivate of a sum, $n$ should be increased by one.

$\displaystyle\frac{1}{1-x}=\sum_{n=0}^\infty x^n \to \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1} \to \frac{2}{(1-x)^3}=\sum_{n=2}^\infty n(n-1)x^{n-2}$

Note 2: You won't see a negative sign because:

$\displaystyle \frac{d}{dx}\left(\frac{1}{(1-x)^2}\right)= (-2)(-1)\frac{1}{(1-x)^3}$

Note 3: This equation is valid only in its correct domain (as a geometric series) as also gimusi said. ($+1$ for that.)

Answer 2

Yes for geometric power series the following hold

• $f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x}\quad |x|<1$

• $f'(x)=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}\quad |x|<1$

• $f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}\quad |x|<1$

we can differentiate and integrate for $|x|<1$.