Question

Find the radius of convergence and interval of convergence of the series. \(\sum\limits_{{\rm{n = 1}}}^\infty {\frac{{{{\rm{2}}^{\rm{n}}}{{{\rm{(x - 2)}}}^{\rm{n}}}}}{{{\rm{(n + 2)!}}}}} \)

Short answer

The sequence is bounded by \({\rm{ - }}\frac{{\rm{1}}}{{\rm{7}}}\)and \(\frac{2}{3}\). Hence it is increasing along with the boundary values.

Step-by-step solution

Finding the Sequence.

To find whether the sequence is increasing or decreasing, check the difference between \({{\rm{a}}_{\rm{n}}}\)and \({{\rm{a}}_{{\rm{n + 1}}}}\)

\(\begin{array}{c}{{\rm{a}}_{\rm{n}}}{\rm{ - }}{{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = }}\frac{{{\rm{2n - 3}}}}{{{\rm{3n + 4}}}}{\rm{ - }}\frac{{{\rm{2(n + 1) - 3}}}}{{{\rm{3(n + 1) + 4}}}}\\{\rm{ = }}\frac{{{\rm{2n - 3}}}}{{{\rm{3n + 4}}}}{\rm{ - }}\frac{{{\rm{2n + 2 - 3}}}}{{{\rm{3n + 3 + 4}}}}\\{\rm{ = }}\frac{{{\rm{2n - 3}}}}{{{\rm{3n + 4}}}}{\rm{ - }}\frac{{{\rm{2n - 1}}}}{{{\rm{3n + 7}}}}\\{\rm{ = }}\frac{{{\rm{(2n - 3)(3n + 7) - (2n - 1)(3n + 4)}}}}{{{\rm{(3n + 4)(3n + 7)}}}}\\{\rm{ = }}\frac{{\left( {{\rm{6}}{{\rm{n}}^{\rm{2}}}{\rm{ + 14n - 9n - 21}}} \right){\rm{ - }}\left( {{\rm{6}}{{\rm{n}}^{\rm{2}}}{\rm{ + 8n - 3n - 4}}} \right)}}{{{\rm{(3n + 4)(3n + 7)}}}}\\{\rm{ = }}\frac{{{\rm{6}}{{\rm{n}}^{\rm{2}}}{\rm{ + 5n - 21 - 6}}{{\rm{n}}^{\rm{2}}}{\rm{ - 5n + 4}}}}{{{\rm{(3n + 4)(3n + 7)}}}}\\{\rm{ = - }}\frac{{{\rm{17}}}}{{{\rm{(3n + 4)(3n + 7)}}}}{\rm{ < 0}}\end{array}\)

Here,\({\rm{3n + 4}}\)and \({\rm{3n + 7}}\)are positive integers.

Therefore,

\(\begin{array}{l}{{\rm{a}}_{\rm{n}}}{\rm{ - }}{{\rm{a}}_{{\rm{n + 1}}}}{\rm{ < 0}}\\{{\rm{a}}_{\rm{n}}}{\rm{ < }}{{\rm{a}}_{{\rm{n + 1}}}}\end{array}\)

Hence it is increasing and it is bounded by its first term.

\(\begin{array}{c}{{\rm{a}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{2 \times 1 - 3}}}}{{{\rm{3 \times 1 + 4}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{7}}}\end{array}\)

Finding the limits.

\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{2n - 3}}}}{{{\rm{3n + 4}}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\frac{{{\rm{2n}}}}{{\rm{n}}}{\rm{ - }}\frac{{\rm{3}}}{{\rm{n}}}}}{{\frac{{{\rm{3n}}}}{{\rm{n}}}{\rm{ + }}\frac{{\rm{4}}}{{\rm{n}}}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{2 - }}\frac{{\rm{3}}}{{\rm{n}}}}}{{{\rm{3 + }}\frac{{\rm{4}}}{{\rm{n}}}}}\\{\rm{ = }}\frac{{\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\rm{2 - }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{3}}}{{\rm{n}}}}}{{\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {\rm{3 + }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{\rm{4}}}{{\rm{n}}}}}\\{\rm{ = }}\frac{{{\rm{2 - 0}}}}{{{\rm{3 - 0}}}}\\{\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}}\\\end{array}\)

Thus, the sequence is bounded by \({\rm{ - }}\frac{{\rm{1}}}{{\rm{7}}}\)and \(\frac{2}{3}\). Hence it is increasing along with the boundary values.