Question

How do you find the power series representation for $\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$ and what is the radius of convergence?

Answer 1

Here, we will first integrate the given function and expand the integrated function using the long division method. Then, on differentiating the expanded expansion, we will obtain the required power series for the function $\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$. For the radius of convergence, we will use the ratio test.

Complete step-by-step answer:
Let us write the function given in the question as
$f\left( x \right) = \dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$……………………$\left( 1 \right)$
Integrating both sides, we get
$\Rightarrow \int {f\left( x \right)dx} = \int {\dfrac{{dx}}{{{{\left( {1 - x} \right)}^2}}}}$
$\Rightarrow \int {f\left( x \right)dx} = \dfrac{{ - {{\left( {1 - x} \right)}^{ - 1}}}}{{ - 1}}$
On simplifying the terms, we get
$\Rightarrow \int {f\left( x \right)dx} = \dfrac{1}{{\left( {1 - x} \right)}}$ ……………………$\left( 2 \right)$
Let us consider the RHS of the above equation as
$g\left( x \right) = \dfrac{1}{{\left( {1 - x} \right)}}$
Now when we divide $1$ by $\left( {1 - x} \right)$, we get
$\dfrac{1}{{\left( {1 - x} \right)}} = 1 + x + {x^2} + {x^3} + ...$
Substituting this in equation $\left( 2 \right)$, we get
$\int {f\left( x \right)dx} = 1 + x + {x^2} + {x^3} + .......$
Differentiating both the sides with respect to $x$, we get
$\dfrac{{d\left( {\int {f\left( x \right)dx} } \right)}}{{dx}} = \dfrac{{d\left( {1 + x + {x^2} + {x^3} + .......} \right)}}{{dx}}$
$\Rightarrow f\left( x \right) = 0 + 1 + 2x + 3{x^2} + .......$
Putting equation $\left( 1 \right)$ in the above equation, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{{{{\left( {1 - x} \right)}^2}}} = 0 + 1 + 2x + 3{x^2} + .......\\\ \Rightarrow \dfrac{1}{{{{\left( {1 - x} \right)}^2}}} = 1{x^{1 - 1}} + 2{x^{2 - 1}} + 3{x^{2 - 1}} + .......\end{array}$
Writing the RHS of the above equation in the contracted form, we finally get
$\Rightarrow \dfrac{1}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 0}^\infty {n{x^{n - 1}}}$………………………..$\left( 3 \right)$
Hence, this the required power series representation for $\dfrac{1}{{{{\left( {1 - x} \right)}^2}}}$.
Now, for the radius of convergence, we have to determine the interval of convergence of the above series. For that, we need to apply the ratio test as below
$L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|$
From $\left( 3 \right)$ we have ${a_n} = n{x^{n - 1}}$. Putting this in above equation, we get
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\left( {n + 1} \right){x^{n + 1}}}}{{n{x^n}}}} \right|$
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{x\left( {n + 1} \right)}}{n}} \right|$
$\Rightarrow L = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n + 1}}{n}} \right|$
As n tends to infinity, $n$ will become close to $n + 1$ and hence $\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n + 1}}{n}} \right| = 1$.
Substituting this in the above equation, we get
$\Rightarrow L = \left| x \right|$
Now, by ratio test, the above limit must be less than one, which means
$\Rightarrow \left| x \right| < 1$
Hence, the radius of convergence is equal to one.

Note:
Here, we need to take proper care of the signs while integrating the function $\dfrac{1}{{\left( {1 - x} \right)}}$. Also, the radius of convergence can be determined using other tests as well such as Raabe’s test, the integral test, etc. But as the ratio test is the easiest of all, we use it to find the radius of convergence. When a power series converges on an interval, then the distance from the center of convergence to the other end of the interval, where it converges, is known as the radius of convergence.