What is the Convergence Radius and Domain of $sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$?












1












$begingroup$


So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$



I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.



It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .



Thanks.










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  • 4




    $begingroup$
    Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
    $endgroup$
    – Wojowu
    Dec 27 '18 at 17:04












  • $begingroup$
    About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:13










  • $begingroup$
    @Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
    $endgroup$
    – Roee
    Dec 27 '18 at 17:56










  • $begingroup$
    @Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
    $endgroup$
    – Roee
    Dec 27 '18 at 17:57










  • $begingroup$
    @Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:58


















1












$begingroup$


So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$



I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.



It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .



Thanks.










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
    $endgroup$
    – Wojowu
    Dec 27 '18 at 17:04












  • $begingroup$
    About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:13










  • $begingroup$
    @Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
    $endgroup$
    – Roee
    Dec 27 '18 at 17:56










  • $begingroup$
    @Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
    $endgroup$
    – Roee
    Dec 27 '18 at 17:57










  • $begingroup$
    @Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:58
















1












1








1





$begingroup$


So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$



I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.



It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .



Thanks.










share|cite|improve this question











$endgroup$




So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$



I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.



It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .



Thanks.







calculus sequences-and-series convergence






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 19:28







Roee

















asked Dec 27 '18 at 17:00









RoeeRoee

374




374








  • 4




    $begingroup$
    Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
    $endgroup$
    – Wojowu
    Dec 27 '18 at 17:04












  • $begingroup$
    About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:13










  • $begingroup$
    @Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
    $endgroup$
    – Roee
    Dec 27 '18 at 17:56










  • $begingroup$
    @Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
    $endgroup$
    – Roee
    Dec 27 '18 at 17:57










  • $begingroup$
    @Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:58
















  • 4




    $begingroup$
    Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
    $endgroup$
    – Wojowu
    Dec 27 '18 at 17:04












  • $begingroup$
    About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:13










  • $begingroup$
    @Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
    $endgroup$
    – Roee
    Dec 27 '18 at 17:56










  • $begingroup$
    @Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
    $endgroup$
    – Roee
    Dec 27 '18 at 17:57










  • $begingroup$
    @Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
    $endgroup$
    – Jakobian
    Dec 27 '18 at 17:58










4




4




$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04






$begingroup$
Hint: $a_n=1/n$ if $n$ is a power of $2$ and $a_n=0$ otherwise. Do you see how to use the formula you mention now?
$endgroup$
– Wojowu
Dec 27 '18 at 17:04














$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13




$begingroup$
About the convergence domain, try Dirichlet test to check it's convergent for every $|z|=1$, with $zneq 1$
$endgroup$
– Jakobian
Dec 27 '18 at 17:13












$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56




$begingroup$
@Wojowu I see, so the partial limits are 0 and 1, the limsup is 1 and the radius is 1, thank you.
$endgroup$
– Roee
Dec 27 '18 at 17:56












$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57




$begingroup$
@Jakobian If I found that the radius is 1, doesn't it mean that the series converges inside the unit circle?
$endgroup$
– Roee
Dec 27 '18 at 17:57












$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58






$begingroup$
@Roee yes, it means that it converges inside the unit circle. The boundary of the circle, it can converge but doesn't have to
$endgroup$
– Jakobian
Dec 27 '18 at 17:58












3 Answers
3






active

oldest

votes


















4












$begingroup$

For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.



The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.



$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$



The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.



For any point $z$ on the unit circle, we have



$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$



The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.



Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.



This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).






share|cite|improve this answer











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  • 2




    $begingroup$
    A typo, Fabry gap theorem instead of Fabry gay theorem
    $endgroup$
    – Jakobian
    Dec 27 '18 at 19:09






  • 1




    $begingroup$
    @Jakobian oops, what a mistake.
    $endgroup$
    – achille hui
    Dec 27 '18 at 19:11



















1












$begingroup$

Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.






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    1












    $begingroup$

    i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.



      The radius of convergence $R$ of the power series
      $$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
      can be computed using the root test.



      $$begin{array}{rll}
      frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
      & color{blue}{leftarrow text{root text}}\
      &= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
      & color{blue}{leftarrow text{definition of "limsup"}}\
      &= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
      &= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
      &= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
      &= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrow text{definition of "limsup" again}}\
      &= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
      &= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
      & color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
      \
      &= 2^0 = 1
      end{array}
      $$



      The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.



      For any point $z$ on the unit circle, we have



      $$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
      = sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$



      The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.



      Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
      There is no way to analytic continue $f(z)$ outside the closed unit disk.



      This means the domain of the power series is exactly the closed unit disk (even if one
      allow analytic continuation).






      share|cite|improve this answer











      $endgroup$









      • 2




        $begingroup$
        A typo, Fabry gap theorem instead of Fabry gay theorem
        $endgroup$
        – Jakobian
        Dec 27 '18 at 19:09






      • 1




        $begingroup$
        @Jakobian oops, what a mistake.
        $endgroup$
        – achille hui
        Dec 27 '18 at 19:11
















      4












      $begingroup$

      For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.



      The radius of convergence $R$ of the power series
      $$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
      can be computed using the root test.



      $$begin{array}{rll}
      frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
      & color{blue}{leftarrow text{root text}}\
      &= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
      & color{blue}{leftarrow text{definition of "limsup"}}\
      &= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
      &= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
      &= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
      &= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrow text{definition of "limsup" again}}\
      &= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
      &= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
      & color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
      \
      &= 2^0 = 1
      end{array}
      $$



      The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.



      For any point $z$ on the unit circle, we have



      $$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
      = sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$



      The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.



      Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
      There is no way to analytic continue $f(z)$ outside the closed unit disk.



      This means the domain of the power series is exactly the closed unit disk (even if one
      allow analytic continuation).






      share|cite|improve this answer











      $endgroup$









      • 2




        $begingroup$
        A typo, Fabry gap theorem instead of Fabry gay theorem
        $endgroup$
        – Jakobian
        Dec 27 '18 at 19:09






      • 1




        $begingroup$
        @Jakobian oops, what a mistake.
        $endgroup$
        – achille hui
        Dec 27 '18 at 19:11














      4












      4








      4





      $begingroup$

      For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.



      The radius of convergence $R$ of the power series
      $$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
      can be computed using the root test.



      $$begin{array}{rll}
      frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
      & color{blue}{leftarrow text{root text}}\
      &= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
      & color{blue}{leftarrow text{definition of "limsup"}}\
      &= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
      &= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
      &= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
      &= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrow text{definition of "limsup" again}}\
      &= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
      &= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
      & color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
      \
      &= 2^0 = 1
      end{array}
      $$



      The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.



      For any point $z$ on the unit circle, we have



      $$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
      = sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$



      The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.



      Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
      There is no way to analytic continue $f(z)$ outside the closed unit disk.



      This means the domain of the power series is exactly the closed unit disk (even if one
      allow analytic continuation).






      share|cite|improve this answer











      $endgroup$



      For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.



      The radius of convergence $R$ of the power series
      $$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
      can be computed using the root test.



      $$begin{array}{rll}
      frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
      & color{blue}{leftarrow text{root text}}\
      &= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
      & color{blue}{leftarrow text{definition of "limsup"}}\
      &= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
      &= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
      &= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
      & color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
      &= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrow text{definition of "limsup" again}}\
      &= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
      & color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
      &= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
      & color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
      \
      &= 2^0 = 1
      end{array}
      $$



      The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.



      For any point $z$ on the unit circle, we have



      $$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
      = sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$



      The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.



      Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
      There is no way to analytic continue $f(z)$ outside the closed unit disk.



      This means the domain of the power series is exactly the closed unit disk (even if one
      allow analytic continuation).







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 27 '18 at 19:10

























      answered Dec 27 '18 at 18:47









      achille huiachille hui

      96.3k5132261




      96.3k5132261








      • 2




        $begingroup$
        A typo, Fabry gap theorem instead of Fabry gay theorem
        $endgroup$
        – Jakobian
        Dec 27 '18 at 19:09






      • 1




        $begingroup$
        @Jakobian oops, what a mistake.
        $endgroup$
        – achille hui
        Dec 27 '18 at 19:11














      • 2




        $begingroup$
        A typo, Fabry gap theorem instead of Fabry gay theorem
        $endgroup$
        – Jakobian
        Dec 27 '18 at 19:09






      • 1




        $begingroup$
        @Jakobian oops, what a mistake.
        $endgroup$
        – achille hui
        Dec 27 '18 at 19:11








      2




      2




      $begingroup$
      A typo, Fabry gap theorem instead of Fabry gay theorem
      $endgroup$
      – Jakobian
      Dec 27 '18 at 19:09




      $begingroup$
      A typo, Fabry gap theorem instead of Fabry gay theorem
      $endgroup$
      – Jakobian
      Dec 27 '18 at 19:09




      1




      1




      $begingroup$
      @Jakobian oops, what a mistake.
      $endgroup$
      – achille hui
      Dec 27 '18 at 19:11




      $begingroup$
      @Jakobian oops, what a mistake.
      $endgroup$
      – achille hui
      Dec 27 '18 at 19:11











      1












      $begingroup$

      Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.






          share|cite|improve this answer









          $endgroup$



          Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 19:48









          ZacharyZachary

          2,3751214




          2,3751214























              1












              $begingroup$

              i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$






                  share|cite|improve this answer









                  $endgroup$



                  i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 27 '18 at 20:08









                  zhw.zhw.

                  74.3k43175




                  74.3k43175






























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