Using the Cauchy-Hadamard Theorem to find a radius of convergence












0















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







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  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40


















0















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question
























  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40
















0












0








0








Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question
















Cauchy-Hadamard Theorem



Consider the formal power series $$f(z) = sum_{n = 0}^{infty}
c_{n}(z - a)^{n} $$



for $a, c_{n} in mathbb{C}$. Then the radius of convergence of $f$
at the point $a$ is given by



$$frac{1}{R} = limsup_{ntoinfty} (|c_{n}|^{1/n}).$$




Can I use this Theorem on this series:



$$sum_{n text{ even}}^{infty} frac{(-1)^{(n - 1)/2} }{prod_{i = 1}^{(n - 1)/2} 2i + 1} cdot x^{n}$$



I tried like this:



$$frac{1}{R} = lim_{ntoinfty} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n}$$



Let the limit equal $L$. Then



$$text{log}(L) = lim_{ntoinfty} frac{1}{n} logleft(frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right) = 0 $$



So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = infty$, though. Does it not work because my sum is only going through even $n$?






sequences-and-series convergence power-series taylor-expansion




sequences-and-series convergence power-series taylor-expansion






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edited Nov 29 '18 at 11:28

























asked Nov 29 '18 at 11:22









joseph

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  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40




















  • $0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
    – mathworker21
    Nov 29 '18 at 11:31












  • I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
    – joseph
    Nov 29 '18 at 11:40


















$0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
– mathworker21
Nov 29 '18 at 11:31






$0^{1/n} = 0$ as well, so you're fine. also, I don't think it's obvious that the limit of $frac{1}{n}log(frac{(-1)^{(n-1)/2}}{prod 2i+1})$ is $0$
– mathworker21
Nov 29 '18 at 11:31














I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
– joseph
Nov 29 '18 at 11:40






I think that the limit is true because the product grows without bound. The numerator only has a $1$ term. But, using the ratio test with $lim_{ntoinfty} |frac{a_{n + 2}}{a_{n}}|$ gave me limit equals $0$, which means convergent for all $x$.
– joseph
Nov 29 '18 at 11:40












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begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
&=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
&=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
&=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
&=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



where I have used Stirling approximation.






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    begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
    &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



    and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
    &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
    &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
    &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



    where I have used Stirling approximation.






    share|cite|improve this answer




























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      begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
      &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



      and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
      &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
      &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
      &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



      where I have used Stirling approximation.






      share|cite|improve this answer


























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        begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
        &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



        and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
        &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
        &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
        &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



        where I have used Stirling approximation.






        share|cite|improve this answer














        begin{align}|c_n| &= begin{cases} left|frac{(-1)^{(n - 1)/2}}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \
        &=begin{cases} left|frac{1}{prod_{i = 1}^{(n - 1)/2} 2i + 1} right|^{1/n} & text{ if } n text{ is even} \0 & text{ if } n text{ is odd}end{cases} \end{align}



        and hence we have begin{align}limsup |c_n| &= lim_{n to infty} left|frac{1}{prod_{i = 1}^{(n - 1)/2} (2i + 1)} right|^{1/n} \
        &=lim_{n to infty} left|frac{n!2^n}{(2n)!} right|^{1/n} \
        &=2 lim_{n to infty} frac{n^frac1{2n}(n/e)}{(2n)^frac1{2n}(2n/e)^2}\
        &=frac{e}2 lim_{n to infty}frac{1}{n }\&= 0end{align}



        where I have used Stirling approximation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '18 at 23:49

























        answered Nov 29 '18 at 12:16









        Siong Thye Goh

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        99.5k1464117






























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