Question

(a) Expand\(f(x) = x/{(1 - x)^2}\) as a power series.

(b) Use part (a) to find the sum of the series \(\sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} \).

Short answer

1. Expand\(f(x) = x/{(1 - x)^2}\) as a power series.

\(\frac{x}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \)

Step-by-step solution

Step 1: Simplify the equation and try to solve

We know that

\(\frac{1}{{{{(1 - x)}^2}}} = \frac{1}{{{{[1 + ( - x)]}^2}}} = {[1 + ( - x)]^{ - 2}}\)

Step 2: Expand

\(\begin{array}{l}\frac{1}{{{{(1 - x)}^2}}} = 1 + ( - 2)( - x) + \frac{{( - 2)( - 3)}}{{2!}}{( - x)^2} + \frac{{( - 2)( - 3)( - 4)}}{{3!}}{( - x)^3}\\ + \frac{{( - 2)( - 3)( - 4)( - 5)}}{{4!}}{( - x)^4} + ......\end{array}\)\(\begin{array}{l}\frac{1}{{{{(1 - x)}^2}}} = 1 + 2x + 3{x^2} + 4{x^3} + 5{x^4} + ......\\\frac{1}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {(n + 1){x^n}} \end{array}\)

Step 3: Further simplification

\(\begin{array}{l}\frac{x}{{{{(1 - x)}^2}}} = x\sum\limits_{n = 1}^\infty {(n + 1){x^n}} \\\frac{x}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {(n + 1){x^{n + 1}}} \\\frac{x}{{{{(1 - x)}^2}}} = x\sum\limits_{n = 1}^\infty {n{x^n}} \end{array}\)

(b) Use part (a) to find the sum of the series \(\sum\limits_{n = 1}^\infty  {\frac{n}{{{2^n}}}} \). \(\sum\limits_{n = 1}^\infty  {\frac{n}{{{2^n}}} = 2} \)Step 1: Given series and simplification

We have the series given as \(S = \sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} = \sum\limits_{n = 1}^\infty n {\left( {\frac{1}{2}} \right)^n}\)

We know that \(\frac{x}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \)

Step 2: Substitution and simplification

Putting \(x = \frac{1}{2}\) we have :

\(\begin{array}{l}\frac{x}{{{{(1 - x)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \\\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{2}} \right)}^n}} = \left[ {\frac{{\left( {\frac{1}{2}} \right)}}{{{{\left( {1 - \left( {\frac{1}{2}} \right)} \right)}^2}}}} \right]\end{array}\)\(\sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} = \left[ {\frac{{\left( {\frac{1}{2}} \right)}}{{{{\left( {\frac{1}{2}} \right)}^2}}}} \right]\)\(\sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} = \left[ {\frac{{\left( {\frac{1}{2}} \right)}}{{\left( {\frac{1}{4}} \right)}}} \right]\)\(\sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} = \frac{{(1)(4)}}{{(2)(1)}}\)\(\sum\limits_{n = 1}^\infty {\frac{n}{{{2^n}}}} = 2\)