Question

Find the Maclaurin series\({\rm{f}}\)and its radius of convergence. You may use either the direct method (definition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for\({{\rm{e}}^{\rm{x}}}\),\({\rm{sinx}}\)and\({\rm{ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{x}}\).

\({\rm{f(x) = }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}\)

Short answer

The series and radius of convergence are \(\sum\limits_{{\rm{n = 0}}}^\infty {{{{\rm{( - 1)}}}^{\rm{n}}}} {{\rm{x}}^{{\rm{n + 2}}}}\),\({\rm{1}}\).

Step-by-step solution

Geometric series rule.

A geometric series is the sum of an infinite number of terms that have a constant ratio between successive terms.

It's important to keep in mind that the total of a geometric series with the initial term \({\rm{a}}\)and the common ratio \({\rm{r}}\)is \(\frac{{\rm{a}}}{{{\rm{1 - r}}}}\).

Compare \(\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}\)with \(\frac{{\rm{a}}}{{{\rm{1 - r}}}}\).

Therefore, can write \(\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}\)

The sum of series with initial term \({\rm{a = }}{{\rm{x}}^{\rm{2}}}\)and common ration \({\rm{r = - x}}\).

\(\begin{array}{c}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{\rm{( - x) + }}{{\rm{x}}^{\rm{2}}}{{\rm{( - x)}}^{\rm{2}}}{\rm{ + }}{{\rm{x}}^{\rm{2}}}{{\rm{( - x)}}^{\rm{3}}}{\rm{ \ldots \ldots }}\\\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{3}}}{\rm{ + }}{{\rm{x}}^{\rm{4}}}{\rm{ - }}{{\rm{x}}^{\rm{5}}}{\rm{ \ldots \ldots }}\\\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}\left( {{\rm{1 - x + }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{3}}}{\rm{ \ldots \ldots }}} \right)\\\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {{{\rm{x}}^{\rm{2}}}} \left( {{{{\rm{( - 1)}}}^{\rm{n}}}{{\rm{x}}^{\rm{n}}}} \right)\\\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{1 + x}}}}{\rm{ = }}\sum\limits_{{\rm{n = 0}}}^\infty {{{{\rm{( - 1)}}}^{\rm{n}}}} {{\rm{x}}^{{\rm{n + 2}}}}\end{array}\)

To find the radius of convergence.

As per the radio test.

\(\sum\limits_{{\rm{n = 0}}}^\infty {{{\rm{a}}_{\rm{n}}}} \)

\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ < 1}}\)

Here,

\({{\rm{a}}_{\rm{n}}}{\rm{ = ( - 1}}{{\rm{)}}^{\rm{n}}}{{\rm{x}}^{{\rm{n + 2}}}}\)

So,

\(\begin{array}{l}{{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = ( - 1}}{{\rm{)}}^{{\rm{n + 1}}}}{{\rm{x}}^{{\rm{n + 3}}}}\\{{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = }}{{\rm{a}}_{\rm{n}}}{\rm{ \times ( - 1)x}}\end{array}\)

Therefore,

\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\rm{| - x|}}\\\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \left| {\frac{{{{\rm{a}}_{{\rm{n + 1}}}}}}{{{{\rm{a}}_{\rm{n}}}}}} \right|{\rm{ = |x| < 1}}\end{array}\)

As a result, the convergence interval is \({\rm{( - 1,1)}}\)and the radius of convergence is \({\rm{1}}\).