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Let $\sum a_n x^n$ be a power series whose radius of convergence is $0<R<\infty$.

What is the radius of convergence, $R'$ of $\sum \frac {n^n}{n!} a_n x^n$?

If I know $\lim_{n\rightarrow\infty} \frac {a_{n+1}}{a_n}$ exists. this would be an easy problem, since (after simplification):

$\frac1{R'}=\lim_{n\rightarrow\infty}(\frac{1+n}n)^n\cdot\frac{a_{n+1}}{a_n}=\frac e R$

But what bothers me is, how can I be sure $\lim \frac{a_{n+1}}{a_n}$ exists in the first place?

edit: $\limsup$ is not enough. For example, in the series $ f_n(x) = \begin{cases} \frac xn, & \text{if $n$ is even} \\ \frac {2x}n, & \text{if $n$ is odd} \end{cases}$

The radius of convergence is 1 but $\limsup \frac {a_{n+1}}{a_n}=2$.

Whyka
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2 Answers2

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Hint: Convergence is not necessary; consider $\limsup$.

Yes
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The most general criterion, which gives an exact formula for the radius in all cases, says that the radius of convergence of a power series is $R$ iff $\frac{1}{R} = \limsup |a_n|^{1/n}$. This is known as Cauchy–Hadamard theorem.

Since you assume that the radius of $\sum a_n x^n$ is $R$, you know that $\limsup |a_n|^{1/n} = \frac{1}{R}$.

Now you want to compute the radius of $\sum \frac{n^n}{n!} a_n x^n$, which is (by the above criterion) 1 over $\limsup |\frac{n^n}{n!}a_n|^{1/n}$. Since $\limsup |\frac{n^n}{n!}|^{1/n} = e$, Arithmetics of limits shows that

$$\limsup |\frac{n^n}{n!}a_n|^{1/n} = \limsup |\frac{n^n}{n!}| \lim |a_n|^{1/n} = e \cdot \frac{1}{R}$$

So the radius is $\frac{R}{e}$.

Appendix: The limit $\limsup |\frac{n^n}{n!}|^{1/n} = e$ can be evaluated as follows: By monotonicity of $\ln x$, $$\int_{1}^{n} \ln x dx < \sum_{i=2}^{n} \ln i < \int_{2}^{n+1} \ln x dx$$ which gives, after division by $n$: $$\frac{n \ln n - n +1}{n} < \frac{\ln n!}{n} < \frac{(n+1)\ln(n+1)-(n+1) - 2 \ln 2 +2}{n}$$ or, phrased differently, $(n!)^{1/n} = \frac{n}{e} \cdot e^{o(1)}$ - which is equivalent to the required limit.

Ofir
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