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Let $$ \sum_{n=0}^\infty c_n (z-a)^n $$ be a power series. If the value $$ r=\underset{n\to\infty}{\lim}\left|\frac{c_n}{c_{n+1}}\right| $$ exists (the limit exists and is a real number), it is the radius of convergence $R$ of the power series. This means that the series converges for all $z$ with $|z-a|<R$.

  1. Is the radius of convergence $R$ the whole $\mathbb{R}$ if the above limit $r$ is $\infty$ (and therefore does not exist in the strict sense)?
  2. Is the radius of convergence $R$ the whole $\mathbb{R}$ if $r'=\underset{n\to\infty}{\lim\operatorname{sup}}\left|\frac{c_n}{c_{n+1}}\right|$ is $\infty$? This is probably not true. What is a counterexample?
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  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. $\endgroup$ – Kasper Apr 26 '13 at 12:56
  • $\begingroup$ @Kasper: why not vote? $\endgroup$ – The Chaz 2.0 Apr 26 '13 at 13:10
  • $\begingroup$ The radius $R$ is an extended real number. It can't be "the whole $\mathbb{R}$". What you mean is that it is $+\infty$. Or that the interval of convergence is "the whole $\mathbb{R}$" $\endgroup$ – Julien Apr 26 '13 at 13:38
  • $\begingroup$ I don't understand the problem with the question. It isn't homework and it is clearly formulated, in my opinion. $\endgroup$ – Ronald Bernard Apr 29 '13 at 13:09
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The answer to question 1. is yes.

Let $R=|z-a|$. Since $\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|=\infty$ there is some $N$ such that $\left|\frac{c_n}{c_{n+1}}\right| > 2R$ for all $n\geq N$. By induction $\left|\frac{c_N}{c_n}\right| > (2R)^{n-N}$.

Thus $$\sum_{n=N}^\infty |c_n (z-a)^n| = |c_N|\sum_{n=N}^\infty\left|\frac{c_n}{c_N}\right|R^{n} < |c_N|\sum_{n=N}^\infty (2R)^{N-n}R^{n} $$ $$= |c_N|R^{N}\sum_{n=0}^\infty 2^{-n} = 2|c_N|R^N$$ so the series converges absolutely.

For a concrete counterexample for 2. take $c_n = \begin{cases} 1 & n\mbox{ even} \\ \frac{1}{n} & n\mbox{odd}\end{cases}$ and note that the series clearly diverges for $|z-a|\geq1$.

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Good question. The answer to 1 is affirmative. For 2, you really need the $\liminf$ to be infinite, which of course implies that the limit exists as $\infty$. This is relevant for finite radii of convergence. More precisely, d'Alembert's 1768 version of the ratio test, as stated on page 70 of Stromberg, states

Let $\sum c_n$ be a series of complex terms with $c_n \neq 0$ for all $n$.

  1. If $\limsup_{n\rightarrow\infty} \left|c_{n+1}/c_n\right|<1$, then the series converges absolutely.
  2. If $\liminf_{n\rightarrow\infty} \left|c_{n+1}/c_n\right|>1$, then hte series diverges.

Upon taking reciprocals in part 1, we see that we need the $\liminf$ for both sides.

To construct a specific counter example to your question, just take $c_{2n}=n^n$ and $c_{2n-1}=n^{-n}$. Then your $\limsup$ is infinite but the series converges nowhere except zero. In fact, you can construct just about any behavior you want using this even/odd alternating definitions.

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