Question

Find the radius of convergence and interval of convergence of the series.\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}{x^n}}}{{2.4.6 \ldots (2n)}}} \)

Short answer

\(R = \infty \) and \(l = ( - \infty ,\infty )\)is the radius of converges and interval of convergence, respectively.

Step-by-step solution

Ratio test

Ratio test

If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L < 1\)then the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)is absolutely convergent.

As given, the series is\(\sum\limits_{n = 1}^\infty {\frac{{{n^2}{x^n}}}{{2.4.6 \ldots (2n)}}} \)

\(\begin{aligned}{\rm{Let }}{a_n} = \frac{{{n^2}{x^n}}}{{2 \cdot 4.6 \ldots (2n)}}\\{a_n} &= \frac{{{n^2}{x^n}}}{{{2^n}n!}}\\{a_n} = \frac{{{n^2}{x^n}}}{{{2^n}(n - 1)!n}}\\{a_n} &= \frac{{n{x^n}}}{{{2^n}(n - 1)!}}\end{aligned}\)

Then,

\({a_{n + 1}} = \frac{{(n + 1){x^{n + 1}}}}{{{2^{n + 1}}((n + 1) - 1)!}}\).

\(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\frac{{\frac{{(n + 1){x^{n + 1}}}}{{{2^{n + 1(n + 1 - 1)!}}}}}}{{\frac{{n{n^n}}}{{2{n^n}(1)!}}}}} \right|\)

Take \(\mathop {\lim }\limits_{n \to \infty } \)on both sides 

Take \(\mathop {\lim }\limits_{n \to \infty } \) on both sides,

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{{{(n + 1)}^{n + 1}}}}{{{2^{n + 1}}(n + 1 - 1)!}}}}{{\frac{{nnn}}{{{2^n}(n - 1)!}}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1){x^{n + 1}}}}{{{2^{n + 1}}(n)!}} \cdot \frac{{{2^n}(n - 1)!}}{{n{x^n}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{(n + 1){x^{n + 1}}}}{{{2^{n + 1}}n(n - 1)!}} \cdot \frac{{{2^n}(n - 1)!}}{{n{x^n}}}} \right|\\\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| &= \mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{2} \cdot \left( {\frac{{n + 1}}{{{n^2}}}} \right)\end{aligned}\)

Simplify the above equation,

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{2} \cdot \left( {\frac{{n + 1}}{{{n^2}}}} \right) &= \frac{{|x|}}{2}\left( {\frac{1}{\infty } + \frac{1}{\infty }} \right)\\\mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{2} \cdot \left( {\frac{{n + 1}}{{{n^2}}}} \right) &= \frac{{|x|}}{2}(0)\\\mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{2} \cdot \left( {\frac{{n + 1}}{{{n^2}}}} \right) &= 0\\\mathop {\lim }\limits_{n \to \infty } \frac{{|x|}}{2} \cdot \left( {\frac{{n + 1}}{{{n^2}}}} \right) < 1\end{aligned}\)

By the Ratio test stated above, the series \(\sum\limits_{n = 1}^\infty {\frac{{{n^2}{x^n}}}{{2.4.6 \ldots (2n)}}} \) is converges for all real\(x\).

Therefore, \(R = \infty \) and \(l = ( - \infty ,\infty )\)is the radius of converges and interval of convergence, respectively.