Convergence of $prod_n=2^inftyleft(1-frac1n^2right)^i^n$, where $i$ is the imaginary unit

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I would like to know an simple way to justify the convergence of the infinite product



$$xi:=prod_n=2^inftyleft(1-frac1n^2right)^i^n,tag1$$
where $i=sqrt-1$ is the imaginary unit.



My question arises when I wondered about a similar question than [1].




Question. Please provide a justification that $xi$ converges*, and the following inequality between its real part and its imaginary part $$Re xi>Im xitag2$$
holds.
Many thanks.




*Only is required to show that the infinite product is well-defined, that is convergent (isn't required a closed-form for $xi$, if my first question is in the literature you can to assume that it converges and solve the second question). My second question arises from a calculation that I did with a CAS.



I think thus that we need to study the partial products of our infinite product and use the perioricity of the sequence $i^n$, as $ngeq 1$ runs over positive integers.



References:



[1] Problem 3673, Crux Mathematicorum, Vol. 38(8), October 2012.










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    up vote
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    down vote

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    I would like to know an simple way to justify the convergence of the infinite product



    $$xi:=prod_n=2^inftyleft(1-frac1n^2right)^i^n,tag1$$
    where $i=sqrt-1$ is the imaginary unit.



    My question arises when I wondered about a similar question than [1].




    Question. Please provide a justification that $xi$ converges*, and the following inequality between its real part and its imaginary part $$Re xi>Im xitag2$$
    holds.
    Many thanks.




    *Only is required to show that the infinite product is well-defined, that is convergent (isn't required a closed-form for $xi$, if my first question is in the literature you can to assume that it converges and solve the second question). My second question arises from a calculation that I did with a CAS.



    I think thus that we need to study the partial products of our infinite product and use the perioricity of the sequence $i^n$, as $ngeq 1$ runs over positive integers.



    References:



    [1] Problem 3673, Crux Mathematicorum, Vol. 38(8), October 2012.










    share|cite|improve this question























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      down vote

      favorite
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      I would like to know an simple way to justify the convergence of the infinite product



      $$xi:=prod_n=2^inftyleft(1-frac1n^2right)^i^n,tag1$$
      where $i=sqrt-1$ is the imaginary unit.



      My question arises when I wondered about a similar question than [1].




      Question. Please provide a justification that $xi$ converges*, and the following inequality between its real part and its imaginary part $$Re xi>Im xitag2$$
      holds.
      Many thanks.




      *Only is required to show that the infinite product is well-defined, that is convergent (isn't required a closed-form for $xi$, if my first question is in the literature you can to assume that it converges and solve the second question). My second question arises from a calculation that I did with a CAS.



      I think thus that we need to study the partial products of our infinite product and use the perioricity of the sequence $i^n$, as $ngeq 1$ runs over positive integers.



      References:



      [1] Problem 3673, Crux Mathematicorum, Vol. 38(8), October 2012.










      share|cite|improve this question













      I would like to know an simple way to justify the convergence of the infinite product



      $$xi:=prod_n=2^inftyleft(1-frac1n^2right)^i^n,tag1$$
      where $i=sqrt-1$ is the imaginary unit.



      My question arises when I wondered about a similar question than [1].




      Question. Please provide a justification that $xi$ converges*, and the following inequality between its real part and its imaginary part $$Re xi>Im xitag2$$
      holds.
      Many thanks.




      *Only is required to show that the infinite product is well-defined, that is convergent (isn't required a closed-form for $xi$, if my first question is in the literature you can to assume that it converges and solve the second question). My second question arises from a calculation that I did with a CAS.



      I think thus that we need to study the partial products of our infinite product and use the perioricity of the sequence $i^n$, as $ngeq 1$ runs over positive integers.



      References:



      [1] Problem 3673, Crux Mathematicorum, Vol. 38(8), October 2012.







      real-analysis complex-analysis reference-request convergence infinite-product






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      asked Jun 25 at 8:03









      user243301

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          2 Answers
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          up vote
          7
          down vote



          accepted










          Rewrite your product using $z^w=e^w log z$ (for positive real $z$).
          $$ prod_n=2^infty left( 1- frac1n^2 right)^i^n =
          prod_n=2^infty e^log left( 1- frac1n^2 right) i^n = e^sum_n=2^infty log left( 1- frac1n^2 right) i^n$$
          This series is absolutely convergent, thus it is convergent. Indeed
          $$sum_n=2^infty left| log left( 1- frac1n^2 right) i^n right| = sum_n=2^inftyleft| log left( 1- frac1n^2 right) right| < infty$$
          is convergent since $log left( 1- frac1n^2 right) sim -frac1n^2$.



          This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.



          In particular, you can separate even terms and odd terms. Recalling that $i^2n= (-1)^n$ and $i^2n+1= (-1)^ni$
          $$prod_n=2^infty left( 1- frac1n^2 right)^i^n = prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n cdot prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni$$
          let's call $K=prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again,
          $$prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni = exp left( sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right) i right) = cos H + i sin H$$
          where
          $$H=sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right)$$



          So you are left to prove that $$cos H > sin H$$
          This follows by the fact that $0 < H <pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations:
          $$log left( 1- frac15^2 right) -log left( 1- frac13^2 right) <H< -log left( 1- frac13^2 right) $$
          i.e.
          $$log(27/25) < H < log(9/8)$$
          And this concludes the proof since $0 < log (27/25)<log (9/8) < pi/4$.






          share|cite|improve this answer






















          • Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
            – user243301
            Jun 25 at 8:48










          • I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
            – zhw.
            Jun 25 at 23:14

















          up vote
          4
          down vote













          This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define



          $$S = sum_n=2^inftyi^nlog (1-1/n^2)=A+iB.$$



          Our infinite product equals $e^A+iB = e^Ae^iB.$ So if suffices to show $0<B<pi/4.$ But with a little work you see



          $$B= ln(9/8) - ln(25/24) + ln(49/48) - cdots$$



          In absolute value these terms are strictly decreasing and $to 0.$ Thus, as is well known, $B$ lies between $0$ and $ln (9/8).$ But since $ln (1+u)<u$ for $u>0,$ we have



          $$ln (9/8) =ln (1+1/8) < 1/8 < pi/4,$$



          and we're done.






          share|cite|improve this answer






















          • Many thank you very much and to the other user answering the question, really, for sharing these calculations.
            – user243301
            Jun 26 at 10:35










          Your 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
          7
          down vote



          accepted










          Rewrite your product using $z^w=e^w log z$ (for positive real $z$).
          $$ prod_n=2^infty left( 1- frac1n^2 right)^i^n =
          prod_n=2^infty e^log left( 1- frac1n^2 right) i^n = e^sum_n=2^infty log left( 1- frac1n^2 right) i^n$$
          This series is absolutely convergent, thus it is convergent. Indeed
          $$sum_n=2^infty left| log left( 1- frac1n^2 right) i^n right| = sum_n=2^inftyleft| log left( 1- frac1n^2 right) right| < infty$$
          is convergent since $log left( 1- frac1n^2 right) sim -frac1n^2$.



          This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.



          In particular, you can separate even terms and odd terms. Recalling that $i^2n= (-1)^n$ and $i^2n+1= (-1)^ni$
          $$prod_n=2^infty left( 1- frac1n^2 right)^i^n = prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n cdot prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni$$
          let's call $K=prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again,
          $$prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni = exp left( sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right) i right) = cos H + i sin H$$
          where
          $$H=sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right)$$



          So you are left to prove that $$cos H > sin H$$
          This follows by the fact that $0 < H <pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations:
          $$log left( 1- frac15^2 right) -log left( 1- frac13^2 right) <H< -log left( 1- frac13^2 right) $$
          i.e.
          $$log(27/25) < H < log(9/8)$$
          And this concludes the proof since $0 < log (27/25)<log (9/8) < pi/4$.






          share|cite|improve this answer






















          • Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
            – user243301
            Jun 25 at 8:48










          • I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
            – zhw.
            Jun 25 at 23:14














          up vote
          7
          down vote



          accepted










          Rewrite your product using $z^w=e^w log z$ (for positive real $z$).
          $$ prod_n=2^infty left( 1- frac1n^2 right)^i^n =
          prod_n=2^infty e^log left( 1- frac1n^2 right) i^n = e^sum_n=2^infty log left( 1- frac1n^2 right) i^n$$
          This series is absolutely convergent, thus it is convergent. Indeed
          $$sum_n=2^infty left| log left( 1- frac1n^2 right) i^n right| = sum_n=2^inftyleft| log left( 1- frac1n^2 right) right| < infty$$
          is convergent since $log left( 1- frac1n^2 right) sim -frac1n^2$.



          This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.



          In particular, you can separate even terms and odd terms. Recalling that $i^2n= (-1)^n$ and $i^2n+1= (-1)^ni$
          $$prod_n=2^infty left( 1- frac1n^2 right)^i^n = prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n cdot prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni$$
          let's call $K=prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again,
          $$prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni = exp left( sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right) i right) = cos H + i sin H$$
          where
          $$H=sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right)$$



          So you are left to prove that $$cos H > sin H$$
          This follows by the fact that $0 < H <pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations:
          $$log left( 1- frac15^2 right) -log left( 1- frac13^2 right) <H< -log left( 1- frac13^2 right) $$
          i.e.
          $$log(27/25) < H < log(9/8)$$
          And this concludes the proof since $0 < log (27/25)<log (9/8) < pi/4$.






          share|cite|improve this answer






















          • Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
            – user243301
            Jun 25 at 8:48










          • I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
            – zhw.
            Jun 25 at 23:14












          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          Rewrite your product using $z^w=e^w log z$ (for positive real $z$).
          $$ prod_n=2^infty left( 1- frac1n^2 right)^i^n =
          prod_n=2^infty e^log left( 1- frac1n^2 right) i^n = e^sum_n=2^infty log left( 1- frac1n^2 right) i^n$$
          This series is absolutely convergent, thus it is convergent. Indeed
          $$sum_n=2^infty left| log left( 1- frac1n^2 right) i^n right| = sum_n=2^inftyleft| log left( 1- frac1n^2 right) right| < infty$$
          is convergent since $log left( 1- frac1n^2 right) sim -frac1n^2$.



          This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.



          In particular, you can separate even terms and odd terms. Recalling that $i^2n= (-1)^n$ and $i^2n+1= (-1)^ni$
          $$prod_n=2^infty left( 1- frac1n^2 right)^i^n = prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n cdot prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni$$
          let's call $K=prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again,
          $$prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni = exp left( sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right) i right) = cos H + i sin H$$
          where
          $$H=sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right)$$



          So you are left to prove that $$cos H > sin H$$
          This follows by the fact that $0 < H <pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations:
          $$log left( 1- frac15^2 right) -log left( 1- frac13^2 right) <H< -log left( 1- frac13^2 right) $$
          i.e.
          $$log(27/25) < H < log(9/8)$$
          And this concludes the proof since $0 < log (27/25)<log (9/8) < pi/4$.






          share|cite|improve this answer














          Rewrite your product using $z^w=e^w log z$ (for positive real $z$).
          $$ prod_n=2^infty left( 1- frac1n^2 right)^i^n =
          prod_n=2^infty e^log left( 1- frac1n^2 right) i^n = e^sum_n=2^infty log left( 1- frac1n^2 right) i^n$$
          This series is absolutely convergent, thus it is convergent. Indeed
          $$sum_n=2^infty left| log left( 1- frac1n^2 right) i^n right| = sum_n=2^inftyleft| log left( 1- frac1n^2 right) right| < infty$$
          is convergent since $log left( 1- frac1n^2 right) sim -frac1n^2$.



          This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.



          In particular, you can separate even terms and odd terms. Recalling that $i^2n= (-1)^n$ and $i^2n+1= (-1)^ni$
          $$prod_n=2^infty left( 1- frac1n^2 right)^i^n = prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n cdot prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni$$
          let's call $K=prod_n=1^inftyleft( 1- frac1(2n)^2 right)^(-1)^n>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again,
          $$prod_n=1^inftyleft( 1- frac1(2n+1)^2 right)^(-1)^ni = exp left( sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right) i right) = cos H + i sin H$$
          where
          $$H=sum_n=1^infty (-1)^n log left( 1- frac1(2n+1)^2 right)$$



          So you are left to prove that $$cos H > sin H$$
          This follows by the fact that $0 < H <pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations:
          $$log left( 1- frac15^2 right) -log left( 1- frac13^2 right) <H< -log left( 1- frac13^2 right) $$
          i.e.
          $$log(27/25) < H < log(9/8)$$
          And this concludes the proof since $0 < log (27/25)<log (9/8) < pi/4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 25 at 9:28

























          answered Jun 25 at 8:20









          Crostul

          26.9k22352




          26.9k22352











          • Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
            – user243301
            Jun 25 at 8:48










          • I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
            – zhw.
            Jun 25 at 23:14
















          • Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
            – user243301
            Jun 25 at 8:48










          • I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
            – zhw.
            Jun 25 at 23:14















          Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
          – user243301
          Jun 25 at 8:48




          Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it).
          – user243301
          Jun 25 at 8:48












          I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
          – zhw.
          Jun 25 at 23:14




          I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer,
          – zhw.
          Jun 25 at 23:14










          up vote
          4
          down vote













          This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define



          $$S = sum_n=2^inftyi^nlog (1-1/n^2)=A+iB.$$



          Our infinite product equals $e^A+iB = e^Ae^iB.$ So if suffices to show $0<B<pi/4.$ But with a little work you see



          $$B= ln(9/8) - ln(25/24) + ln(49/48) - cdots$$



          In absolute value these terms are strictly decreasing and $to 0.$ Thus, as is well known, $B$ lies between $0$ and $ln (9/8).$ But since $ln (1+u)<u$ for $u>0,$ we have



          $$ln (9/8) =ln (1+1/8) < 1/8 < pi/4,$$



          and we're done.






          share|cite|improve this answer






















          • Many thank you very much and to the other user answering the question, really, for sharing these calculations.
            – user243301
            Jun 26 at 10:35














          up vote
          4
          down vote













          This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define



          $$S = sum_n=2^inftyi^nlog (1-1/n^2)=A+iB.$$



          Our infinite product equals $e^A+iB = e^Ae^iB.$ So if suffices to show $0<B<pi/4.$ But with a little work you see



          $$B= ln(9/8) - ln(25/24) + ln(49/48) - cdots$$



          In absolute value these terms are strictly decreasing and $to 0.$ Thus, as is well known, $B$ lies between $0$ and $ln (9/8).$ But since $ln (1+u)<u$ for $u>0,$ we have



          $$ln (9/8) =ln (1+1/8) < 1/8 < pi/4,$$



          and we're done.






          share|cite|improve this answer






















          • Many thank you very much and to the other user answering the question, really, for sharing these calculations.
            – user243301
            Jun 26 at 10:35












          up vote
          4
          down vote










          up vote
          4
          down vote









          This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define



          $$S = sum_n=2^inftyi^nlog (1-1/n^2)=A+iB.$$



          Our infinite product equals $e^A+iB = e^Ae^iB.$ So if suffices to show $0<B<pi/4.$ But with a little work you see



          $$B= ln(9/8) - ln(25/24) + ln(49/48) - cdots$$



          In absolute value these terms are strictly decreasing and $to 0.$ Thus, as is well known, $B$ lies between $0$ and $ln (9/8).$ But since $ln (1+u)<u$ for $u>0,$ we have



          $$ln (9/8) =ln (1+1/8) < 1/8 < pi/4,$$



          and we're done.






          share|cite|improve this answer














          This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define



          $$S = sum_n=2^inftyi^nlog (1-1/n^2)=A+iB.$$



          Our infinite product equals $e^A+iB = e^Ae^iB.$ So if suffices to show $0<B<pi/4.$ But with a little work you see



          $$B= ln(9/8) - ln(25/24) + ln(49/48) - cdots$$



          In absolute value these terms are strictly decreasing and $to 0.$ Thus, as is well known, $B$ lies between $0$ and $ln (9/8).$ But since $ln (1+u)<u$ for $u>0,$ we have



          $$ln (9/8) =ln (1+1/8) < 1/8 < pi/4,$$



          and we're done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 25 at 23:13

























          answered Jun 25 at 18:52









          zhw.

          67.8k42872




          67.8k42872











          • Many thank you very much and to the other user answering the question, really, for sharing these calculations.
            – user243301
            Jun 26 at 10:35
















          • Many thank you very much and to the other user answering the question, really, for sharing these calculations.
            – user243301
            Jun 26 at 10:35















          Many thank you very much and to the other user answering the question, really, for sharing these calculations.
          – user243301
          Jun 26 at 10:35




          Many thank you very much and to the other user answering the question, really, for sharing these calculations.
          – user243301
          Jun 26 at 10:35

















           

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