Reduce this series

Question

Let $$f(x)= \sum_{n=2}^\infty nx^n4^n$$ How do we reduce this? I know that $$\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$ and $$\sum_{n=0}^\infty a^n x^n = \frac{1}{1-ax}$$

But how do I combine both?

Answer 1

i have the partial sum and it looks like the following ${\frac { \left( 4\,x \right) ^{m+1} \left( 4\, \left( m+1 \right) x-m- 1-4\,x \right) }{ \left( 4\,x-1 \right) ^{2}}}-16\,{\frac {{x}^{2} \left( 4\,x-2 \right) }{ \left( 4\,x-1 \right) ^{2}}} $

Answer 2

Here's a way to find a closed expression for this sum

\begin{align*} f(x)&=\sum_{n=2}^{\infty}nx^n4^n\\ &=\sum_{n=2}^{\infty}n(4x)^n\\ &=4x\sum_{n=2}^{\infty}n(4x)^{n-1}\\ &=x\frac{d}{dx}\sum_{n=2}^{\infty}(4x)^{n}\tag{1}\\ &=x\left(\frac{d}{dx}\left(\frac{1}{1-4x}-1-4x\right)\right)\tag{2}\\ &=x\left(\frac{4}{(1-4x)^2}-4\right)\tag{3}\\ &=4x\left(\frac{1}{(1-4x)^2}-1\right) \end{align*}

Observe that

• in (1) we use $\frac{d}{dx}(4x)^n=4n(4x)^{n-1}$

• in (2) we use the geometric series $\sum_{n=0}^{\infty}(4x)^n=\frac{1}{1-4x}$

• in (3) we differentiate $\frac{1}{1-4x}$ to get $\frac{4}{(1-4x)^2}$

Answer 3

You already know $$ \sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{1} $$ (the $n=0$ term is $0$)

Just substitute $x\mapsto4x$ in $(1)$ to get $$ \sum_{n=1}^\infty n(4x)^n=\frac{4x}{(1-4x)^2}\tag{2} $$ then subtract the $n=1$ term from both sides $$ \sum_{n=2}^\infty n4^nx^n=\frac{4x}{(1-4x)^2}-4x\tag{3} $$ You can simplify $(3)$ however seems best

Note that since $(1)$ converges for $|x|\lt1$, $(3)$ converges for $|4x|\lt1$, or $|x|\lt\frac14$.