text: Radius of convergence of power series which has factorial term
  
  I am trying to find radius of convergence of the following power series: $\sum_{n\geq 1} n^n z^{n!}$
  
  I tried ratio test but it became complicated, I have never seen such radius of convergence problem with factorial.
  
  • The ratio test can also be applied in this series, using consecutive nonvanishing terms. – hardmath Aug 12 '14 at 13:47
  
  By the Cauchy-Hadamard formula; the radius of convergence of a power series isgiven by $$\frac1{R}=\limsup |a_n|^{1/n}$$
  
  Now , we may plug in and see that $|n^n|^{1/n!} \to 1$ as $n\to \infty$ .So we conclude that $R=1$.
  
  • hmm, I probably miss something, but Cauchy-Hadamard formula requires $a_n = c_n (z-z_0)^n$ format, while we have $n!$. – Alex Aug 12 '14 at 12:20
  • @Alex The coefficients are $0$ for all $n$ that aren't factorials. – Daniel Fischer Aug 12 '14 at 12:21
  • @DanielFischer $a_n(z)=n^nz^{n!}$, and we want find the value of $z$ for which the series $\sum a_n(z)$ converge, in this case $\sqrt[n]{|a_n(z)|}\neq |n|^{\frac{1}{n!}}$. – Hamou Aug 12 '14 at 12:47
  • @Hamou For the Cauchy-Hadamard formula, you consider only the coefficients $a_n$ of the power series $$\sum_{n=0}^\infty a_n (z-z_0)^n.$$ Here, we have $$a_n = \begin{cases} k^k &, n = k!\\ 0 &, (\forall k)(n \neq k!).\end{cases}$$ And thus $$\limsup_{n\to\infty} \lvert a_n\rvert^{1/n} = \lim_{k\to\infty} \lvert k^k\rvert^{1/k!} = \lim_{k\to\infty} k^{1/(k-1)!} = 1.$$ – Daniel Fischer Aug 12 '14 at 12:52
  • @DanielFischer: you should have put this as an answer! – Alex Aug 12 '14 at 14:50
  
  By Cauchy's criterion we have $R=1$. EDIT $\alpha_n=\sqrt[n]{n^n|z|^{n!}}=n|z|^{(n-1)!}$, if $|z|>1$ it is clear that $\alpha_n\to \infty$ (the series diverge). If $|z|<1$, then $\alpha_n=\left[\dfrac{n}{(n-1)!}\right]\left[(n-1)!|z|^{(n-1)!}\right]\to 0$ (the series converge).
  
  • you mean not by ratio test or root test? – user169306 Aug 12 '14 at 11:14
  • Sorry, root test, but by French it called "Critère de Cauchy", and I am not very good in English. Any correction is welcome – Hamou Aug 12 '14 at 11:19
  • The ratio test is also known as Cauchy's ratio test, so additional information is necessary to disambiguate the meaning. – hardmath Aug 12 '14 at 11:36
  • The last answer is edited – Hamou Aug 12 '14 at 11:56
explanation: The document contains a math question about finding the radius of convergence of a specific power series, and it also includes an answer using the Cauchy-Hadamard formula. The answer is further clarified in the comments.
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distilabel_metadata.content: Radius of convergence of power series which has factorial term
  
  I am trying to find radius of convergence of the following power series: $\sum_{n\geq 1} n^n z^{n!}$
  
  I tried ratio test but it became complicated, I have never seen such radius of convergence problem with factorial.
  
  • The ratio test can also be applied in this series, using consecutive nonvanishing terms. – hardmath Aug 12 '14 at 13:47
  
  By the Cauchy-Hadamard formula; the radius of convergence of a power series isgiven by $$\frac1{R}=\limsup |a_n|^{1/n}$$
  
  Now , we may plug in and see that $|n^n|^{1/n!} \to 1$ as $n\to \infty$ .So we conclude that $R=1$.
  
  • hmm, I probably miss something, but Cauchy-Hadamard formula requires $a_n = c_n (z-z_0)^n$ format, while we have $n!$. – Alex Aug 12 '14 at 12:20
  • @Alex The coefficients are $0$ for all $n$ that aren't factorials. – Daniel Fischer Aug 12 '14 at 12:21
  • @DanielFischer $a_n(z)=n^nz^{n!}$, and we want find the value of $z$ for which the series $\sum a_n(z)$ converge, in this case $\sqrt[n]{|a_n(z)|}\neq |n|^{\frac{1}{n!}}$. – Hamou Aug 12 '14 at 12:47
  • @Hamou For the Cauchy-Hadamard formula, you consider only the coefficients $a_n$ of the power series $$\sum_{n=0}^\infty a_n (z-z_0)^n.$$ Here, we have $$a_n = \begin{cases} k^k &, n = k!\\ 0 &, (\forall k)(n \neq k!).\end{cases}$$ And thus $$\limsup_{n\to\infty} \lvert a_n\rvert^{1/n} = \lim_{k\to\infty} \lvert k^k\rvert^{1/k!} = \lim_{k\to\infty} k^{1/(k-1)!} = 1.$$ – Daniel Fischer Aug 12 '14 at 12:52
  • @DanielFischer: you should have put this as an answer! – Alex Aug 12 '14 at 14:50
  
  By Cauchy's criterion we have $R=1$. EDIT $\alpha_n=\sqrt[n]{n^n|z|^{n!}}=n|z|^{(n-1)!}$, if $|z|>1$ it is clear that $\alpha_n\to \infty$ (the series diverge). If $|z|<1$, then $\alpha_n=\left[\dfrac{n}{(n-1)!}\right]\left[(n-1)!|z|^{(n-1)!}\right]\to 0$ (the series converge).
  
  • you mean not by ratio test or root test? – user169306 Aug 12 '14 at 11:14
  • Sorry, root test, but by French it called "Critère de Cauchy", and I am not very good in English. Any correction is welcome – Hamou Aug 12 '14 at 11:19
  • The ratio test is also known as Cauchy's ratio test, so additional information is necessary to disambiguate the meaning. – hardmath Aug 12 '14 at 11:36
  • The last answer is edited – Hamou Aug 12 '14 at 11:56
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distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question about finding the radius of convergence of a specific power series, and it also includes an answer using the Cauchy-Hadamard formula. The answer is further clarified in the comments.
  
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