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Radius of convergence of power series $$\sum _{n=1} ^{\infty} \frac{n!}{n^n} x^{2n}$$

I have calculated $\overline\lim a_n^{\frac{1}{n}}= \infty $.

So radius of convergence should be zero. But textbook answer is $\sqrt{e}$.

is my answer wrong ? if yes please solve it for $\overline\lim a_n^{\frac{1}{n}} $.

Thanks in advanced

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    $\begingroup$ You already had excellent answer but you can notice that the series converges for $x=1$ (since $0<\frac{n!}{n^n}<\frac{1}{n(n-1)}$) so the radius is at least $1$. $\endgroup$ – Taladris Jun 7 at 13:02
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Since$$\sqrt[n]{\frac{n!}{n^n}}=\frac{\sqrt[n]{n!}}n$$and since, by Stirling's approximation,$$\sqrt[n]{n!}\approx\sqrt[n]{\sqrt{2\pi n}}\frac ne,$$you have$$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}=\frac1e$$and therefore the radius of convergence is $\sqrt e$. It would be $e$ if you had $x^n$ in your power series, but you have $x^{2n}(=(x^2)^n)$.

But if you want to avoid Stirling's approximation, then I strongly suggest that you use the ratio test.

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Since $$ \sqrt{2\pi}n^{n+\frac12}e^{-n}\leq n!\leq en^{n+\frac12}e^{-n} $$ we have \begin{align*} \limsup_n \left[\frac{n!}{n^n}x^{2n}\right]^{1/n} &=x^2\limsup_n \frac{(n!)^{1/n}}n\\ &=x^2\limsup_n \frac{[Cn^{n+\frac12}e^{-n}]^{1/n}}n\\ &=\frac{x^2}{e} \end{align*} so series converges if $\lvert x^2\rvert<e$ and diverges if $\lvert x^2\rvert>e$, i.e., radius of convergence is $\sqrt{e}$.

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$$\dfrac{1}{R}=\lim_{n \to \infty} | \dfrac {a_{n+1}}{a_{n}}|= \lim_{n \to \infty} | \dfrac{(n+1)! n^{n}}{(n+1)^{n+1} n!}| = \big(\dfrac{1}{1+ \frac{1}{n}}\big)^{n} = \dfrac{1}{e}$$ so the series converges for $|x^2|<e$ i.e. $|x| < \sqrt{e}$

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Let me give another answer, another way to get $\lim_{n \to \infty} a_n$ since there is a common trick to get such limit:

$$ \lim_{n \to \infty} \sqrt[n]{\frac{n!}{n^n}} = \lim_{n \to \infty} \sqrt[n]{\frac{n(n-1)(n-2) \cdots 1}{n\cdot n \cdot n \cdots \cdot n}} = e^{\ln\bigg( \lim_{n \to \infty} \sqrt[n]{\frac{n(n-1)(n-2) \cdots 1}{n\cdot n \cdot n \cdots \cdot n}} \bigg) } = e^{\bigg( \lim_{n \to \infty} \frac{1}{n}\ln \frac{n(n-1)(n-2) \cdots 1}{n\cdot n \cdot n \cdots \cdot n} \bigg) } = e^{\bigg( \lim_{n \to \infty} \frac{1}{n}\bigg( \ln \frac{n}{n} + \ln \frac{n-1}{n} + \cdots + \ln\frac{1}{n} \bigg) \bigg) } = e^{\bigg( \lim_{n \to \infty} \frac{1}{n}\sum_i^n \ln\frac{i}{n} \bigg) } = e^{\int_0^1 \ln(x) dx } = e^{-1} = \frac{1}{e}$$

This is the type of problem where the trick is to convert the limit to a Riemann sum, then from the Riemann sum you can use the Fundamental Theorem of Calculus to help you to get the answer! Hopefully this help you to see it another way!

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