Science:Math Exam Resources/Courses/MATH105/April 2014/Question 02 (b)/Solution 1

Let $a_k=\frac{x^k}{10^{k+1}(k+1)!}$ be the terms in the series. From the ratio test, we are guaranteed absolute convergence when $\underset{k\to \mathrm{\infty }}{\lim }|\frac{a_{k+1}}{a_k}|<1$.

With $a_k=\frac{x^k}{10^{k+1}(k+1)!}$, we have $a_{k+1}=\frac{x^{k+1}}{10^{k+2}(k+2)!}$ and thus

$\begin{array}{rl}|\frac{a_{k+1}}{a_k}|& =|\frac{\frac{x^{k+1}}{10^{k+2}(k+2)!}}{\frac{x^k}{10^{k+1}(k+1)!}}|\\ & =|\frac{x^{k+1}10^{k+1}(k+1)!}{x^{k}10^{k+2}(k+2)!}|\\ & =|\frac{x}{10}\frac{(k+1)!}{(k+2)!}|.\end{array}$

We recall that $k!=k(k-1)(k-2)...(3)(2)(1)$ and thus in general $k!=k(k-1)!$. This also means that $(k+2)!=(k+2)(k+1)!$. Thus,

$\begin{array}{rl}|\frac{a_{k+1}}{a_k}|& =|\frac{x(k+1)!}{10(k+2)(k+1)!}|\\ & =|\frac{x}{10(k+2)}|\end{array}$.

Computing $\underset{k\to \mathrm{\infty }}{\lim }|\frac{x}{10(k+2)}|=0<1$ for all x. Hence, the series converges absolutely for all x-values and the radius of convergence is $\mathrm{\infty }$.