# Find a power series representation for the function and determine the radius of

Find a power series representation for the function and determine the radius of convergence. $f\left(x\right)=\frac{x}{{\left(1+4x\right)}^{2}}$
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Faiza Fuller

Step 1
We will start with the power series of $\frac{1}{1-x}$
We know that
$\frac{1}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{x}^{n}$
Step 1 : Replace x with $-4x$
$\frac{1}{1-\left(-4x\right)}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4x\right)}^{n}$
$\frac{1}{1+4x}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}{x}^{n}$
Step 2
Differentiate both sides
$\frac{d}{dx}\left[\frac{1}{1+4x}\right]=\frac{d}{dx}\left[\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}{x}^{n}\right]$
$-\frac{4}{{\left(1+4x\right)}^{2}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-4\right)}^{n}\frac{d\left[{x}^{n}\right]}{dx}$
$-\frac{4}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-4\right)}^{n}\left(n\right){x}^{n-1}$
Not that the sum in RHS is starting from $n=1$, because the first term died during differentiation
Divide both sides by $-4$, To get
$\frac{1}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-4\right)}^{n-1}\left(n\right){x}^{n-1}$
$\frac{1}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n-1}\cdot {4}^{n-1}\cdot \left(n\right){x}^{n-1}$
Step 3 : Multiply both sides by x
$\frac{x}{{\left(1+4x\right)}^{2}}=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n-1}{\left(4\right)}^{n-1}\left(n\right){x}^{n}$
If you want you can start the sum from $n=0$, but if you do that, you should replace n with $n+1$ in the sum