I'm studying for my analysis final and we were given a bunch of practice questions about the radius of convergence. I understand how to apply the ratio test to find the radius of convergence, but I'm not really sure how to show the following:

Show that the radius of convergence R of the power series $\sum a_nx^n$ is given by lim$|\frac{a_{n}}{a_n+1}|$ whenever it exists.

We're also supposed to find:

Give an example of a power series where this limit does not exist.

Thanks in advance.


Think about what the Ratio Test says. Given a power series $$\sum_{n=0}^\infty a_n x^n$$ the ratio test says that the series converges absolutely if $$L = \lim_{n\to\infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| |x| < 1$$ (it may or may not converge absolutely if that is $1$, but this won't affect the radius of convergence) Equivalently, we have that $$|x| < \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$ And since $R$ is defined to be the radius of the largest disk (interval) in which the series converges absolutely, this tells us that $$R = \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$ and if this limit exists and is non-zero, this tells us in particular that the radius of convergence $R$ is finite, and in this case, is given by precisely $$R = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$

Concerning the example, try $$\sum_{n=0}^\infty n!\cdot x^n$$ In this case, $a_n = n!$, and $R = \lim_{n\to\infty} \left|\frac{n!}{(n+1)!}\right| = 0$, but $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ does not exist, but this yields $R = 0$, and the series converges only at $0$.

If instead, you want a series where the limit $\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|$ doesn't exist, take $$\sum_{n=0}^\infty \frac{x^n}{n!}$$ i.e., $a_n = \frac{1}{n!}$, and in this case, $$\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n\to\infty} \left|\frac{(n+1)!}{n!}\right| = \lim_{n\to\infty} (n+1)$$ which fails to exist, but does in fact yield $R = \infty$, since $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 0$ is the reciprocal of the radius of convergence.

  • $\begingroup$ If $a_n=n!$ then $$\frac{a_n}{a_{n+1}}=\frac{1}{n+1}$$ converges to $0$. Assuming the user meant to write the reciprocal of what was written. So disregard this comment. :) $\endgroup$
    – user142299
    Apr 20 '14 at 3:28
  • $\begingroup$ Thanks for the suggestion. I went ahead and clarified my answer to reflect your comment, giving an example where $\lim_{n\to\infty} \frac{a_n}{a_{n+1}}$ fails to exist, and also where $\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}$ fails to exist, and clarified what $R$ would be in each of these cases. $\endgroup$ Apr 20 '14 at 17:51

I think you wrote the radius down incorrectly. The ratio test says that the series will converge absolutely if $$\lim_{n\to\infty}{\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|}=|x|\lim_{n\to\infty}{\left|{a_{n+1}\over a_n}\right|}<1$$ which happens when $$|x|<\lim{\left|a_n\over a_{n+1}\right|}$$ so the radius is that. I'm not sure what level of rigour your class is looking for, but hopefully that helps.

For the second part, you could take $a_n=0$, or less trivially $a_n=\sin n$.

  • $\begingroup$ (Disregard comment - I had a typo). $\endgroup$
    – user114014
    Apr 20 '14 at 3:35
  • $\begingroup$ @user114014 Well I and the other poster and wikipedia all say it is the way I have it, so I guess it's either a mistake or perhaps you're misreading something? $\endgroup$
    – user142299
    Apr 20 '14 at 3:37
  • $\begingroup$ Do those examples for the second part apply for the correct R (lim a_n / a_n+1) or the one I originally wrote (reciprocal)? $\endgroup$
    – user114014
    Apr 20 '14 at 3:42
  • $\begingroup$ @user114014 Both, actually. Also look at the example in the other poster's response, that one applies for your original writing only. $\endgroup$
    – user142299
    Apr 20 '14 at 3:43
  • $\begingroup$ I'm also confused because I was following the examples on this website (sosmath.com/calculus/radcon/radcon02/radcon02.html) and there they use the original ratio test. So which one is it - which is used to find R? Thanks. $\endgroup$
    – user114014
    Apr 20 '14 at 3:47

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