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

MATH105 April 2014

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Question 02 (b)
Find the radius of convergence of the power series $\displaystyle \sum _{k=0}^{\infty }{\frac {x^{k}}{10^{k+1}(k+1)!}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that the radius of convergence for a series can be found by determining the range of x-values where the series converges absolutely. The ratio test will give you information about absolute convergence.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\infty$.