How to evaluate $ lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n $ where $i=sqrt{-1}$?












0












$begingroup$



How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.




I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?






sequences-and-series limits complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's very hard to understand what you wrote. Try mathjax
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 21:38












  • $begingroup$
    Express $(1+i)/sqrt{2}=exp(ipi/4)$.
    $endgroup$
    – Diger
    Nov 24 '18 at 21:39








  • 1




    $begingroup$
    L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:43






  • 1




    $begingroup$
    If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:46












  • $begingroup$
    Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 21:57
















0












$begingroup$



How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.




I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?






sequences-and-series limits complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's very hard to understand what you wrote. Try mathjax
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 21:38












  • $begingroup$
    Express $(1+i)/sqrt{2}=exp(ipi/4)$.
    $endgroup$
    – Diger
    Nov 24 '18 at 21:39








  • 1




    $begingroup$
    L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:43






  • 1




    $begingroup$
    If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:46












  • $begingroup$
    Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 21:57














0












0








0





$begingroup$



How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.




I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?






sequences-and-series limits complex-numbers







share|cite|improve this question











$endgroup$





How to evaluate the following limit?
$$
lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
$$
Here $i=sqrt{-1}$.




I got:
$$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n
= lim_{nto infty} frac{(i-1)^n}{n(sqrt{2})^n}
$$
I know the lower part goes to infinity but what to do with the upper part? Is that usefull to use squeeze theorem or is there any simplier way?






sequences-and-series limits complex-numbers




sequences-and-series limits complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 23:16









user587192

1,789315




1,789315










asked Nov 24 '18 at 21:35









L. SpyL. Spy

12




12








  • 1




    $begingroup$
    It's very hard to understand what you wrote. Try mathjax
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 21:38












  • $begingroup$
    Express $(1+i)/sqrt{2}=exp(ipi/4)$.
    $endgroup$
    – Diger
    Nov 24 '18 at 21:39








  • 1




    $begingroup$
    L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:43






  • 1




    $begingroup$
    If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:46












  • $begingroup$
    Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 21:57














  • 1




    $begingroup$
    It's very hard to understand what you wrote. Try mathjax
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 21:38












  • $begingroup$
    Express $(1+i)/sqrt{2}=exp(ipi/4)$.
    $endgroup$
    – Diger
    Nov 24 '18 at 21:39








  • 1




    $begingroup$
    L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:43






  • 1




    $begingroup$
    If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
    $endgroup$
    – amWhy
    Nov 24 '18 at 21:46












  • $begingroup$
    Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
    $endgroup$
    – Shubham Johri
    Nov 24 '18 at 21:57








1




1




$begingroup$
It's very hard to understand what you wrote. Try mathjax
$endgroup$
– DonAntonio
Nov 24 '18 at 21:38






$begingroup$
It's very hard to understand what you wrote. Try mathjax
$endgroup$
– DonAntonio
Nov 24 '18 at 21:38














$begingroup$
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
$endgroup$
– Diger
Nov 24 '18 at 21:39






$begingroup$
Express $(1+i)/sqrt{2}=exp(ipi/4)$.
$endgroup$
– Diger
Nov 24 '18 at 21:39






1




1




$begingroup$
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
$endgroup$
– amWhy
Nov 24 '18 at 21:43




$begingroup$
L.Spy Talk of "mathjax" might be confusing to you. Sorry about that. Mathjax is a way to format mathematical expressions so they render very nicely, like you'd see in a textbook. Here is a really handy tutorial for learning mathjax. Anything you want under a square root sign, you can format as $sqrt{blah blah}$. A limit of a function f(x) from $n to infty$ can be written $lim_{nto infty} f(x)$. Just a few pointers.
$endgroup$
– amWhy
Nov 24 '18 at 21:43




1




1




$begingroup$
If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
$endgroup$
– amWhy
Nov 24 '18 at 21:46






$begingroup$
If your function $f(x)$ is a fraction, you can write it as follows $f(x) = frac{"numerator here"}{"denominator here"}$
$endgroup$
– amWhy
Nov 24 '18 at 21:46














$begingroup$
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
$endgroup$
– Shubham Johri
Nov 24 '18 at 21:57




$begingroup$
Write $frac{1+i}{sqrt{2}}=e^{ipi/4}implies (frac{1+i}{sqrt{2}})^n=e^{i npi/4}$
$endgroup$
– Shubham Johri
Nov 24 '18 at 21:57










2 Answers
2






active

oldest

votes


















4












$begingroup$

We have
$$
left|frac{1+i}{sqrt{2}}right|=1
$$
Thus
$$
left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
    $endgroup$
    – L. Spy
    Nov 25 '18 at 9:25










  • $begingroup$
    @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
    $endgroup$
    – egreg
    Nov 25 '18 at 9:40





















1












$begingroup$

$$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$






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









    4












    $begingroup$

    We have
    $$
    left|frac{1+i}{sqrt{2}}right|=1
    $$
    Thus
    $$
    left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
      $endgroup$
      – L. Spy
      Nov 25 '18 at 9:25










    • $begingroup$
      @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
      $endgroup$
      – egreg
      Nov 25 '18 at 9:40


















    4












    $begingroup$

    We have
    $$
    left|frac{1+i}{sqrt{2}}right|=1
    $$
    Thus
    $$
    left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
      $endgroup$
      – L. Spy
      Nov 25 '18 at 9:25










    • $begingroup$
      @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
      $endgroup$
      – egreg
      Nov 25 '18 at 9:40
















    4












    4








    4





    $begingroup$

    We have
    $$
    left|frac{1+i}{sqrt{2}}right|=1
    $$
    Thus
    $$
    left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
    $$






    share|cite|improve this answer









    $endgroup$



    We have
    $$
    left|frac{1+i}{sqrt{2}}right|=1
    $$
    Thus
    $$
    left|frac{i}{n}left(frac{1+i}{sqrt{2}}right)^{!n}right|=frac{1}{n}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 '18 at 22:53









    egregegreg

    180k1485202




    180k1485202












    • $begingroup$
      egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
      $endgroup$
      – L. Spy
      Nov 25 '18 at 9:25










    • $begingroup$
      @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
      $endgroup$
      – egreg
      Nov 25 '18 at 9:40




















    • $begingroup$
      egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
      $endgroup$
      – L. Spy
      Nov 25 '18 at 9:25










    • $begingroup$
      @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
      $endgroup$
      – egreg
      Nov 25 '18 at 9:40


















    $begingroup$
    egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
    $endgroup$
    – L. Spy
    Nov 25 '18 at 9:25




    $begingroup$
    egreg-How it could be 1? I thought that (1+i)/√2 is one of the roots of i=(a+bi)^2
    $endgroup$
    – L. Spy
    Nov 25 '18 at 9:25












    $begingroup$
    @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
    $endgroup$
    – egreg
    Nov 25 '18 at 9:40






    $begingroup$
    @L.Spy $sqrt{(1/sqrt{2})^2+(1/sqrt{2})^2}=sqrt{1/2+1/2}=1$. With your (more complicated) approach: since $|i|=1$, also its square roots have modulus $1$.
    $endgroup$
    – egreg
    Nov 25 '18 at 9:40













    1












    $begingroup$

    $$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$






        share|cite|improve this answer









        $endgroup$



        $$lim_{nto infty} |frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n|=lim_{nto infty} {1over n}=0$$therefore $$lim_{nto infty} frac{i}{n}left(frac{1+i}{sqrt{2}}right)^n=0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 0:31









        Mostafa AyazMostafa Ayaz

        15.2k3939




        15.2k3939






























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