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Let $\sum_{n\ge 0} a_nz^n$ be a power series with $a_i\in\mathbb{C}$.

Fact 1. The limit $\lim\sup |a_n|^{1/n}$ always exists. If it is finite and non-zero, its reciprocal is the radicus of convergence of the series.

Fact 2. If the limit $\lim_{n\rightarrow \infty} |a_{n+1}|/|a_n|$ exists and non-zero, then its reciprocal is radius of convergence.

Question: Let $\sum a_nz^n$ be a power series with following conditions:

  • all $a_i$ are non-zero.

  • The radius of convergence is finite and non-zero (say $R=2$).

Is it necessary that $\lim_{n\rightarrow \infty} |a_{n+1}|/|a_n|$ exists?


This may be obvious/silly question; but I came across it with example of $\sum z^{n!}$. Although many coefficients are zero, and the limit of rations of coefficients not exists, the radius of convergence is $1$; I am simply considering a case where all coefficients are non-zero.

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  • $\begingroup$ Well, doesn't the last example answer your question? the limit of the quotient doesn't exist, yet the lim sup of the $\;n\,-$th root exists and equals one... $\endgroup$ – DonAntonio Feb 27 '18 at 8:53
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    $\begingroup$ One can also find counterexamples, where $a_n \neq 0$: Define $a_n = \begin{cases} n^{-2} & \text{if } n \text{ is odd} \\ n^{-3} & \text{otherwise} \end{cases}$. Here $\liminf_{n \rightarrow \infty} a_{n+1}/a_n =0$ and $\limsup_{n \rightarrow \infty} a_{n+1}/a_n = \infty$. $\endgroup$ – p4sch Feb 27 '18 at 8:59
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Take $$f(x)=\dfrac{1}{1-x^2}+\dfrac{1}{1-x}=2+x+2x^2+x^3+2x^4+x^5+2x^6+...\qquad ,\qquad |x|<1$$ therefore $$a_n=2 \qquad ,\qquad\text{n is even}\\a_n=1 \qquad ,\qquad\text{n is odd}$$so the limit doesn't exist and the radius of convergence is $1$.

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  • $\begingroup$ Sorry I'm gonna fix it....... $\endgroup$ – Mostafa Ayaz Feb 27 '18 at 9:21
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Take $a_n=n$ if $n\not=k!$ for all $k$ and $a_n=1$ for $n=k!$. For the corresponding series the sequence of $\vert a_{n+1}/a_n\vert$ has subsequences converging to 0 and $\infty$ respectively.

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