Find the convergence radius of series











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I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?






limits convergence complex-numbers







share|cite|improve this question
























  • Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
    – Clement C.
    Nov 19 at 18:24










  • isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
    – user3132457
    Nov 19 at 18:26












  • (also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
    – Clement C.
    Nov 19 at 18:26












  • @OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
    – Clement C.
    Nov 19 at 18:27






  • 1




    So it should be $${frac{e^n + e^{-n}}{2}}$$?
    – user3132457
    Nov 19 at 18:28

















up vote
0
down vote

favorite












I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?






limits convergence complex-numbers







share|cite|improve this question
























  • Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
    – Clement C.
    Nov 19 at 18:24










  • isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
    – user3132457
    Nov 19 at 18:26












  • (also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
    – Clement C.
    Nov 19 at 18:26












  • @OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
    – Clement C.
    Nov 19 at 18:27






  • 1




    So it should be $${frac{e^n + e^{-n}}{2}}$$?
    – user3132457
    Nov 19 at 18:28















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?






limits convergence complex-numbers







share|cite|improve this question















I need to find the convergence radius of the following series:
$$sum_{n=0}^infty cos in times z^n$$
Since
$$cos in = cosh n = {frac{e^{in} + e^{-in}}{2}}$$
therefore, using D'Alembert's property I find the limit as
$$lim {frac{a_{n+1}}{a_n}} = {frac{e^{i(n+1)} + e^{-i(n+1)}}{2}} times z^n times z times {frac{2}{(e^{in} + e^{-in})z^n}} = lim z times {frac{e^{i(n+1)} + e^{-i(n+1)}}{e^{in} + e^{-in}}} = z$$
Therefore the radius should be 1. But the answer on the book is ${frac{1}{e}}$.
What's wrong with my solution?






limits convergence complex-numbers




limits convergence complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 18:26









Bernard

117k637109




117k637109










asked Nov 19 at 18:21









user3132457

856




856












  • Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
    – Clement C.
    Nov 19 at 18:24










  • isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
    – user3132457
    Nov 19 at 18:26












  • (also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
    – Clement C.
    Nov 19 at 18:26












  • @OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
    – Clement C.
    Nov 19 at 18:27






  • 1




    So it should be $${frac{e^n + e^{-n}}{2}}$$?
    – user3132457
    Nov 19 at 18:28




















  • Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
    – Clement C.
    Nov 19 at 18:24










  • isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
    – user3132457
    Nov 19 at 18:26












  • (also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
    – Clement C.
    Nov 19 at 18:26












  • @OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
    – Clement C.
    Nov 19 at 18:27






  • 1




    So it should be $${frac{e^n + e^{-n}}{2}}$$?
    – user3132457
    Nov 19 at 18:28


















Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24




Are you sure this is the definition of $cosh n$ you wrote? Looks like $cos n$...
– Clement C.
Nov 19 at 18:24












isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26






isn't it the same as $cos n$, only with $i$s added to the exponentiation of $e$?
– user3132457
Nov 19 at 18:26














(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26






(also, even assuming it's correct, not clear how you compute your last limit. The ratio does not tend to $1$, whether you look at what you wrote or the correct one.)
– Clement C.
Nov 19 at 18:26














@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27




@OP: $$cos x = frac{e^{ix}+e^{-ix}}{2}$$ and $$cosh x = frac{e^{x}+e^{-x}}{2}$$ you picked the wrong one.
– Clement C.
Nov 19 at 18:27




1




1




So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28






So it should be $${frac{e^n + e^{-n}}{2}}$$?
– user3132457
Nov 19 at 18:28












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










You have two errors here. The first is that
$$
cosh n = frac{e^n + e^{-n}}{2}
$$
not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
$$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).



So let's compute (1):
$$
frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
= frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
= frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
$$
Therefore,
$$
lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
$$
which explains why the radius is $frac{1}{e}$.






share|cite|improve this answer




























    up vote
    1
    down vote













    Hint:



    We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence



    far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$






    share|cite|improve this answer























    • The OP is using the ratio test.
      – Clement C.
      Nov 19 at 18:52










    • @ClementC, Please find updated answer
      – lab bhattacharjee
      Nov 19 at 19:05











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    up vote
    1
    down vote



    accepted










    You have two errors here. The first is that
    $$
    cosh n = frac{e^n + e^{-n}}{2}
    $$
    not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
    $$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).



    So let's compute (1):
    $$
    frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
    = frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
    = frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
    $$
    Therefore,
    $$
    lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
    $$
    which explains why the radius is $frac{1}{e}$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      You have two errors here. The first is that
      $$
      cosh n = frac{e^n + e^{-n}}{2}
      $$
      not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
      $$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).



      So let's compute (1):
      $$
      frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
      = frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
      = frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
      $$
      Therefore,
      $$
      lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
      $$
      which explains why the radius is $frac{1}{e}$.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        You have two errors here. The first is that
        $$
        cosh n = frac{e^n + e^{-n}}{2}
        $$
        not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
        $$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).



        So let's compute (1):
        $$
        frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
        = frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
        = frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
        $$
        Therefore,
        $$
        lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
        $$
        which explains why the radius is $frac{1}{e}$.






        share|cite|improve this answer












        You have two errors here. The first is that
        $$
        cosh n = frac{e^n + e^{-n}}{2}
        $$
        not what you wrote (which is $cos n$. To remember: $cosh$ is unbounded on $mathbb{R}$, so you have "true" exponentials, while $cos$ is bounded, so you have $e^{i x}$ which is bounded). So you need to compute
        $$lim_{ntoinfty} frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}} tag{1}$$ and not $lim_{ntoinfty} frac{e^{i(n+1)}+e^{-i(n+1)}}{e^{in}+e^{-in}}$ (where you also made a mistake, the second: this limit does not exist, anyway!).



        So let's compute (1):
        $$
        frac{e^{n+1}+e^{-(n+1)}}{e^{n}+e^{-n}}
        = frac{e^n(e+e^{-(2n+1)})}{e^n(1+e^{-2n})} =
        = frac{e+e^{-(2n+1)}}{1+e^{-2n}} xrightarrow[ntoinfty]{} frac{e+0}{1+0} = e,.
        $$
        Therefore,
        $$
        lim_{ntoinfty} frac{a_{n+1}}{a_n} = zcdot e tag{2}
        $$
        which explains why the radius is $frac{1}{e}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 18:36









        Clement C.

        49.2k33785




        49.2k33785






















            up vote
            1
            down vote













            Hint:



            We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence



            far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$






            share|cite|improve this answer























            • The OP is using the ratio test.
              – Clement C.
              Nov 19 at 18:52










            • @ClementC, Please find updated answer
              – lab bhattacharjee
              Nov 19 at 19:05















            up vote
            1
            down vote













            Hint:



            We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence



            far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$






            share|cite|improve this answer























            • The OP is using the ratio test.
              – Clement C.
              Nov 19 at 18:52










            • @ClementC, Please find updated answer
              – lab bhattacharjee
              Nov 19 at 19:05













            up vote
            1
            down vote










            up vote
            1
            down vote









            Hint:



            We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence



            far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$






            share|cite|improve this answer














            Hint:



            We can use https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence



            far more easily,so that we need $$|ez|<1$$ and $$left|dfrac zeright|<1$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 19 at 19:04

























            answered Nov 19 at 18:51









            lab bhattacharjee

            222k15155273




            222k15155273












            • The OP is using the ratio test.
              – Clement C.
              Nov 19 at 18:52










            • @ClementC, Please find updated answer
              – lab bhattacharjee
              Nov 19 at 19:05


















            • The OP is using the ratio test.
              – Clement C.
              Nov 19 at 18:52










            • @ClementC, Please find updated answer
              – lab bhattacharjee
              Nov 19 at 19:05
















            The OP is using the ratio test.
            – Clement C.
            Nov 19 at 18:52




            The OP is using the ratio test.
            – Clement C.
            Nov 19 at 18:52












            @ClementC, Please find updated answer
            – lab bhattacharjee
            Nov 19 at 19:05




            @ClementC, Please find updated answer
            – lab bhattacharjee
            Nov 19 at 19:05


















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