Question

Find the Maclaurin series for \(f\) and its radius of the convergence.

The \(n\)th derivative of the function \(f(x)\) at the point 0 is, \({f^n}(0) = (n + 1)!\) for\(n = 1,2,3, \cdots \).

Short answer

The series is \(f(x) = \sum\limits_{n = 0}^\infty {(n + 1)} {x^n}\) and radius of the convergence is 1.

Step-by-step solution

Concept used

(1) The expansion of the Maclaurin series\(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(0)}}{{n!}}} \)is,

\(f(0) + \frac{{{f^\prime }0}}{{1!}}x + \frac{{{f^{\prime \prime }}(0)}}{{2!}}{x^2} + \frac{{{f^{\prime \prime \prime }}(0)}}{{3!}}{x^3} + \cdots (1)\)

(2) The Ratio Test:

(i) 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 unconditionally convergent (and it is also convergent).

(ii) If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L > 1\)or\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \infty \), the series\(\sum\limits_{n = 1}^\infty {{a_n}} \)then becomes divergent.

(iii) If\(\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 1\), the Ratio Test inconclusive; that is, no conclusion can be drawn about the convergence or divergence of\(\sum\limits_{n = 1}^\infty {{a_n}} \).

Find the Maclaurin series

As given, the \(n\)th derivative of the function \(f(x)\) at the point 0 is, \({f^n}(0) = (n + 1)!\) for \(n = 1,2,3, \cdots \).

Consider the given term \({f^n}(0) = (n + 1)!(2)\)

Substitute\({f^n}(0) = (n + 1)!{\mathop{\rm in}\nolimits} f(x) = \sum\limits_{n = 0}^\infty {\frac{{{f^{(n)}}(0)}}{{n!}}} \)

\(\begin{aligned}{l}f(x) &= \sum\limits_{n = 0}^\infty {\frac{{(n + 1)!}}{{n!}}} {x^n}\\f(x) &= \sum\limits_{n = 0}^\infty {\frac{{(n + 1)n!}}{{n!}}} {x^n}\\f(x) &= \sum\limits_{n = 0}^\infty {(n + 1)} {x^n}\end{aligned}\)

Hence, the Maclaurin series for \(f(x)\) is\(\sum\limits_{n = 0}^\infty {(n + 1)} {x^n}\).

Here, the \(n\)th term is\({a_n} = (n + 1){x^n}\)

Then, \({a_{n + 1}} = (n + 2){x^{n + 1}}\)

Obtain the limit

Obtain the limit \(\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

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

Thus, by the above concept of 2 (i), the series is convergence when\(|x|\angle 1\).

Therefore, it can be concluded that given series is converges with radius 1.

Hence, the radius of convergence is 1.