So I just did my analysis exam and I had this question where I needed to find the convergence radius for this power series. I did not really know what to do because of the $n!$ and the $(-1)^n.$ There must be a trick since it was part of a longer question. Any ideas? $$\sum _{n=0}^{\infty }\frac{\left(-1\right)^{n+1}\left(n-1\right)}{n!}\left(x-1\right)^n$$
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Hint: the ratio test handles factorials well. – Sean Roberson Aug 07 '17 at 18:13
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But doesn't the sequence need to not be $0$ for all $n$? – Aug 07 '17 at 18:14
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@MolecularPrime Why would it need to be always nonzero? – Xander Henderson Aug 07 '17 at 18:17
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$\mathrm{:and:}\lim {n\to \infty }|\frac{a{n+1}}{a_n}|$, how am I to divide by $a_n$ if it's $0$? – Aug 07 '17 at 18:18
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@MolecularPrime But that is a limit. If $a_n$ is eventually nonzero, why would you need $a_n$ to be nonzero for all $n$? You are looking at limiting behaviour, not what happens when $n$ is small. – Xander Henderson Aug 07 '17 at 18:20
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I agree. I thought I could remove the terms that are $0$ but my stupid book said to only use that when $a_n$ is not $0$ for all $n$. – Aug 07 '17 at 18:21
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No coefficient is zero. – Sean Roberson Aug 07 '17 at 18:22
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1@SeanRoberson Except when $n=1$. ;) – Xander Henderson Aug 07 '17 at 18:22
2 Answers
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The radius of convergence of a power series $$ \sum_{n=0}^{\infty} a_n (x-x_0)^n $$ is given by $$ \frac{1}{\limsup_{n\to\infty} \sqrt[n]{\left|a_n\right|}}. $$ This can be checking using the root test for series convergence. But $$ \limsup_{n\to\infty} \sqrt[n]{ \left| \frac{(-1)^{n+1} (n-1)}{n!} \right|} = \limsup_{n\to\infty} \sqrt[n]{\frac{n-1}{n!}} = 0. $$ From this, it follows that the radius of convergence is $\infty$. That is, the series converges absolutely for all $x$.
Xander Henderson
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Just use the ratio test - an AP calculus classic trick.
\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right | &= \lim_{n \to \infty} \left| \frac{(-1)^{n+2} (n+1-1) (x-1)^{n+1} }{(n+1)!} \cdot \frac{ n!}{(-1)^{n+1} (n-1) (x-1)^{n} } \right| \\ &= \lim_{n \to \infty} \left| \frac{(-1) (n) (x-1)}{(n+1)} \cdot \frac{1}{(n-1)} \right| \\ &= |x-1| \lim_{n \to \infty} \left| \frac{1}{2} \left(\frac{1}{n + 1} + \frac{1}{n-1} \right) \right| \end{align*}
You can finish it here - or use Xander's answer (I was beat while writing this up).
Sean Roberson
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