Studying the convergence of $sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$ The Next CEO of...

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Studying the convergence of $sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$



The Next CEO of Stack OverflowConvergence of $sumlimits_{n=2}^infty frac{1}{n^alpha ln^beta (n)} $ for nonnegative $alpha$ and $beta$Test for convergence $sum_{n=1}^{infty} frac{1}{2^sqrt{n}}$Studying the absolute and conditional convergence of $sum_{n=1}^{infty} frac{(-1)^n}{n^a(ln(n))^3}$Does the series $sum_limits{n=1}^{infty}frac{z^n}{a^{sqrt{n}}}$ converge for $a<1$?Convergence using Weierstrass's test $sum_limits{n=1}^{infty}frac{n!}{a^{n^2}}z^n$Study the convergence of $sum_limits{n=1}^{infty}frac{(-1)^n}{(x+n)^p}$Convergence domain of $sum_limits{n=1}^{infty}(-1)^{n+1}frac{1}{n^x}$?Absolute convergence of $sum_limits{n=1}^{infty}(-1)^{n+1}e^{-nsin x}$Find the convergence domain of the series $sum_limits{n=1}^{infty}frac{(-1)^{n-1}}{n3^n(x-5)^n}$Studying the convergence of $sum_limits{n=1}^{infty}frac{cos(in)}{2^n}$












2












$begingroup$



Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$




I thought of applying the root test:



$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$



However is $e^i>1$?



Questions:



1) Is the application of the root test right? What conclusion can I take?



2) Which would be other alternative ways of solving the question?



Thanks in advance!






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
    $endgroup$
    – Conrad
    Mar 17 at 17:10










  • $begingroup$
    Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
    $endgroup$
    – xbh
    Mar 17 at 17:21






  • 1




    $begingroup$
    @Conrad The root test DOES apply, but is inconclusive.
    $endgroup$
    – Mark Viola
    Mar 17 at 17:44










  • $begingroup$
    ???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
    $endgroup$
    – Conrad
    Mar 17 at 18:15


















2












$begingroup$



Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$




I thought of applying the root test:



$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$



However is $e^i>1$?



Questions:



1) Is the application of the root test right? What conclusion can I take?



2) Which would be other alternative ways of solving the question?



Thanks in advance!






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
    $endgroup$
    – Conrad
    Mar 17 at 17:10










  • $begingroup$
    Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
    $endgroup$
    – xbh
    Mar 17 at 17:21






  • 1




    $begingroup$
    @Conrad The root test DOES apply, but is inconclusive.
    $endgroup$
    – Mark Viola
    Mar 17 at 17:44










  • $begingroup$
    ???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
    $endgroup$
    – Conrad
    Mar 17 at 18:15
















2












2








2





$begingroup$



Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$




I thought of applying the root test:



$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$



However is $e^i>1$?



Questions:



1) Is the application of the root test right? What conclusion can I take?



2) Which would be other alternative ways of solving the question?



Thanks in advance!






sequences-and-series complex-analysis







share|cite|improve this question









$endgroup$





Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$




I thought of applying the root test:



$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$



However is $e^i>1$?



Questions:



1) Is the application of the root test right? What conclusion can I take?



2) Which would be other alternative ways of solving the question?



Thanks in advance!






sequences-and-series complex-analysis




sequences-and-series complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 17 at 17:05









Pedro GomesPedro Gomes

1,9822721




1,9822721












  • $begingroup$
    No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
    $endgroup$
    – Conrad
    Mar 17 at 17:10










  • $begingroup$
    Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
    $endgroup$
    – xbh
    Mar 17 at 17:21






  • 1




    $begingroup$
    @Conrad The root test DOES apply, but is inconclusive.
    $endgroup$
    – Mark Viola
    Mar 17 at 17:44










  • $begingroup$
    ???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
    $endgroup$
    – Conrad
    Mar 17 at 18:15




















  • $begingroup$
    No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
    $endgroup$
    – Conrad
    Mar 17 at 17:10










  • $begingroup$
    Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
    $endgroup$
    – xbh
    Mar 17 at 17:21






  • 1




    $begingroup$
    @Conrad The root test DOES apply, but is inconclusive.
    $endgroup$
    – Mark Viola
    Mar 17 at 17:44










  • $begingroup$
    ???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
    $endgroup$
    – Conrad
    Mar 17 at 18:15


















$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10




$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10












$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21




$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21




1




1




$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44




$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44












$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15






$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15












3 Answers
3






active

oldest

votes


















4












$begingroup$

When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.



On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Application of the ROOT TEST shows



    $$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$



    So, the root test is inconclusive.



    On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).



    The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
      $endgroup$
      – Mark Viola
      Mar 25 at 17:55



















    2












    $begingroup$

    As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.



    Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.



    EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.






    share|cite|improve this answer











    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.



      On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.



        On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.



          On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.






          share|cite|improve this answer











          $endgroup$



          When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.



          On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 17 at 17:38

























          answered Mar 17 at 17:10









          José Carlos SantosJosé Carlos Santos

          171k23132240




          171k23132240























              2












              $begingroup$

              Application of the ROOT TEST shows



              $$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$



              So, the root test is inconclusive.



              On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).



              The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Mar 25 at 17:55
















              2












              $begingroup$

              Application of the ROOT TEST shows



              $$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$



              So, the root test is inconclusive.



              On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).



              The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Mar 25 at 17:55














              2












              2








              2





              $begingroup$

              Application of the ROOT TEST shows



              $$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$



              So, the root test is inconclusive.



              On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).



              The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!






              share|cite|improve this answer









              $endgroup$



              Application of the ROOT TEST shows



              $$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$



              So, the root test is inconclusive.



              On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).



              The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 17 at 17:41









              Mark ViolaMark Viola

              134k1278176




              134k1278176












              • $begingroup$
                @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Mar 25 at 17:55


















              • $begingroup$
                @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
                $endgroup$
                – Mark Viola
                Mar 25 at 17:55
















              $begingroup$
              @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              Mar 25 at 17:55




              $begingroup$
              @pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              Mar 25 at 17:55











              2












              $begingroup$

              As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.



              Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.



              EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.



                Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.



                EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.



                  Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.



                  EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.






                  share|cite|improve this answer











                  $endgroup$



                  As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.



                  Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.



                  EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 17 at 17:58

























                  answered Mar 17 at 17:46









                  Ryan GouldenRyan Goulden

                  495310




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