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 $\lim _{k\rightarrow \infty }|{\frac {a_{k+1}}{a_{k}}}|<1$.

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

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

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,

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

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