Find the range of convergence of the series$,,sum_{n=0}^infty {frac{z^n}{1+z^{2n}}}$











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The series I have is
$$displaystylesum_{n=0}^infty {dfrac{z^n}{1+z^{2n}}}$$



The same series with absolute values is:
$$displaystylesum_{n=0}^infty {dfrac{|z|^n}{1+|z|^{2n}}}$$



Using D'Alembert's principle,
$$displaystylelim {dfrac{a_{n+1}}{a_n}} = {dfrac{|z|^n cdot |z|}{1+|z|^{2n} cdot |z|}} cdot {dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$



The convergence range is when $|z| < 1$. But the book answer is $|z| ne 1$.










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  • What's $Z_n$? Is it actually $Z^n$?
    – xbh
    Nov 18 at 9:33










  • Yes, sorry. Updated
    – user3132457
    Nov 18 at 9:34










  • I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
    – xbh
    Nov 18 at 9:36










  • So what is wrong?
    – user3132457
    Nov 18 at 9:38










  • Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
    – xbh
    Nov 18 at 9:41

















up vote
1
down vote

favorite
1












The series I have is
$$displaystylesum_{n=0}^infty {dfrac{z^n}{1+z^{2n}}}$$



The same series with absolute values is:
$$displaystylesum_{n=0}^infty {dfrac{|z|^n}{1+|z|^{2n}}}$$



Using D'Alembert's principle,
$$displaystylelim {dfrac{a_{n+1}}{a_n}} = {dfrac{|z|^n cdot |z|}{1+|z|^{2n} cdot |z|}} cdot {dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$



The convergence range is when $|z| < 1$. But the book answer is $|z| ne 1$.










share|cite|improve this question
























  • What's $Z_n$? Is it actually $Z^n$?
    – xbh
    Nov 18 at 9:33










  • Yes, sorry. Updated
    – user3132457
    Nov 18 at 9:34










  • I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
    – xbh
    Nov 18 at 9:36










  • So what is wrong?
    – user3132457
    Nov 18 at 9:38










  • Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
    – xbh
    Nov 18 at 9:41















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





The series I have is
$$displaystylesum_{n=0}^infty {dfrac{z^n}{1+z^{2n}}}$$



The same series with absolute values is:
$$displaystylesum_{n=0}^infty {dfrac{|z|^n}{1+|z|^{2n}}}$$



Using D'Alembert's principle,
$$displaystylelim {dfrac{a_{n+1}}{a_n}} = {dfrac{|z|^n cdot |z|}{1+|z|^{2n} cdot |z|}} cdot {dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$



The convergence range is when $|z| < 1$. But the book answer is $|z| ne 1$.










share|cite|improve this question















The series I have is
$$displaystylesum_{n=0}^infty {dfrac{z^n}{1+z^{2n}}}$$



The same series with absolute values is:
$$displaystylesum_{n=0}^infty {dfrac{|z|^n}{1+|z|^{2n}}}$$



Using D'Alembert's principle,
$$displaystylelim {dfrac{a_{n+1}}{a_n}} = {dfrac{|z|^n cdot |z|}{1+|z|^{2n} cdot |z|}} cdot {dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$



The convergence range is when $|z| < 1$. But the book answer is $|z| ne 1$.







sequences-and-series complex-analysis convergence absolute-convergence






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share|cite|improve this question













share|cite|improve this question




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edited Nov 18 at 10:02









Yiorgos S. Smyrlis

62.1k1383162




62.1k1383162










asked Nov 18 at 9:29









user3132457

856




856












  • What's $Z_n$? Is it actually $Z^n$?
    – xbh
    Nov 18 at 9:33










  • Yes, sorry. Updated
    – user3132457
    Nov 18 at 9:34










  • I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
    – xbh
    Nov 18 at 9:36










  • So what is wrong?
    – user3132457
    Nov 18 at 9:38










  • Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
    – xbh
    Nov 18 at 9:41




















  • What's $Z_n$? Is it actually $Z^n$?
    – xbh
    Nov 18 at 9:33










  • Yes, sorry. Updated
    – user3132457
    Nov 18 at 9:34










  • I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
    – xbh
    Nov 18 at 9:36










  • So what is wrong?
    – user3132457
    Nov 18 at 9:38










  • Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
    – xbh
    Nov 18 at 9:41


















What's $Z_n$? Is it actually $Z^n$?
– xbh
Nov 18 at 9:33




What's $Z_n$? Is it actually $Z^n$?
– xbh
Nov 18 at 9:33












Yes, sorry. Updated
– user3132457
Nov 18 at 9:34




Yes, sorry. Updated
– user3132457
Nov 18 at 9:34












I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
– xbh
Nov 18 at 9:36




I don't think the calculation of $lim a_{n+1}/a_n$ is correct.
– xbh
Nov 18 at 9:36












So what is wrong?
– user3132457
Nov 18 at 9:38




So what is wrong?
– user3132457
Nov 18 at 9:38












Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
– xbh
Nov 18 at 9:41






Wouldn't $|z|>1$ and $|z|<1$ lead to different results?
– xbh
Nov 18 at 9:41












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










If $|z|=r<1$, and $nge 1$, then
$$
left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}=frac{r^n}{1-r^{2n}}
<frac{r^n}{1-r}
$$

and hence the series
$$
sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
$$

converges, due to Comparison Test.



If $|z|=1$, and in particular $z=i$, then the series is not even definable.



Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $zinmathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=exp(ik/2^ell)$ are singularities (not isolated) of the series, for all $k,ellinmathbb N$.






share|cite|improve this answer























  • I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
    – user3132457
    Nov 18 at 9:56










  • This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
    – Yiorgos S. Smyrlis
    Nov 18 at 10:00










  • I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
    – user3132457
    Nov 18 at 10:02




















up vote
0
down vote













$$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
=frac{z^n}{z^{2n}+1}\
=a_n(z)$$

So it converges for $z$ whenever it converges for $1/z$.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    If $|z|=r<1$, and $nge 1$, then
    $$
    left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}=frac{r^n}{1-r^{2n}}
    <frac{r^n}{1-r}
    $$

    and hence the series
    $$
    sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
    $$

    converges, due to Comparison Test.



    If $|z|=1$, and in particular $z=i$, then the series is not even definable.



    Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $zinmathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=exp(ik/2^ell)$ are singularities (not isolated) of the series, for all $k,ellinmathbb N$.






    share|cite|improve this answer























    • I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
      – user3132457
      Nov 18 at 9:56










    • This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
      – Yiorgos S. Smyrlis
      Nov 18 at 10:00










    • I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
      – user3132457
      Nov 18 at 10:02

















    up vote
    1
    down vote



    accepted










    If $|z|=r<1$, and $nge 1$, then
    $$
    left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}=frac{r^n}{1-r^{2n}}
    <frac{r^n}{1-r}
    $$

    and hence the series
    $$
    sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
    $$

    converges, due to Comparison Test.



    If $|z|=1$, and in particular $z=i$, then the series is not even definable.



    Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $zinmathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=exp(ik/2^ell)$ are singularities (not isolated) of the series, for all $k,ellinmathbb N$.






    share|cite|improve this answer























    • I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
      – user3132457
      Nov 18 at 9:56










    • This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
      – Yiorgos S. Smyrlis
      Nov 18 at 10:00










    • I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
      – user3132457
      Nov 18 at 10:02















    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    If $|z|=r<1$, and $nge 1$, then
    $$
    left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}=frac{r^n}{1-r^{2n}}
    <frac{r^n}{1-r}
    $$

    and hence the series
    $$
    sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
    $$

    converges, due to Comparison Test.



    If $|z|=1$, and in particular $z=i$, then the series is not even definable.



    Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $zinmathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=exp(ik/2^ell)$ are singularities (not isolated) of the series, for all $k,ellinmathbb N$.






    share|cite|improve this answer














    If $|z|=r<1$, and $nge 1$, then
    $$
    left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}=frac{r^n}{1-r^{2n}}
    <frac{r^n}{1-r}
    $$

    and hence the series
    $$
    sum_{n=0}^inftyfrac{z^n}{1+z^{2n}}
    $$

    converges, due to Comparison Test.



    If $|z|=1$, and in particular $z=i$, then the series is not even definable.



    Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $zinmathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=exp(ik/2^ell)$ are singularities (not isolated) of the series, for all $k,ellinmathbb N$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 18 at 10:09

























    answered Nov 18 at 9:53









    Yiorgos S. Smyrlis

    62.1k1383162




    62.1k1383162












    • I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
      – user3132457
      Nov 18 at 9:56










    • This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
      – Yiorgos S. Smyrlis
      Nov 18 at 10:00










    • I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
      – user3132457
      Nov 18 at 10:02




















    • I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
      – user3132457
      Nov 18 at 9:56










    • This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
      – Yiorgos S. Smyrlis
      Nov 18 at 10:00










    • I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
      – user3132457
      Nov 18 at 10:02


















    I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
    – user3132457
    Nov 18 at 9:56




    I don't get how $$left|frac{z^n}{1+z^{2n}}right|le frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold.
    – user3132457
    Nov 18 at 9:56












    This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
    – Yiorgos S. Smyrlis
    Nov 18 at 10:00




    This is true when $|z|=r<1$, as mentioned in the beginning of the answer.
    – Yiorgos S. Smyrlis
    Nov 18 at 10:00












    I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
    – user3132457
    Nov 18 at 10:02






    I know, but the answer is $|z| ne 1$ which means $|z| > 1$ is possible to have.
    – user3132457
    Nov 18 at 10:02












    up vote
    0
    down vote













    $$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
    =frac{z^n}{z^{2n}+1}\
    =a_n(z)$$

    So it converges for $z$ whenever it converges for $1/z$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      $$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
      =frac{z^n}{z^{2n}+1}\
      =a_n(z)$$

      So it converges for $z$ whenever it converges for $1/z$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        $$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
        =frac{z^n}{z^{2n}+1}\
        =a_n(z)$$

        So it converges for $z$ whenever it converges for $1/z$.






        share|cite|improve this answer












        $$a_nleft(frac1zright)=frac{1/z^n}{1+1/z^{2n}}\
        =frac{z^n}{z^{2n}+1}\
        =a_n(z)$$

        So it converges for $z$ whenever it converges for $1/z$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 18 at 10:38









        Empy2

        33.3k12261




        33.3k12261






























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