Finding this Power Series's Interval of Convergence












0












$begingroup$


We consider the power series $sum_{n=1}^{infty} a_nx^n$ where $a_n = frac{2^k}{k}$ if $n$ is even, and $0$ otherwise.



I understand how to find that this series has radius of convergence $R = frac{1}{sqrt{2}}$. What I don't understand is how to now find its interval of convergence. I know I'm supposed to plug $R$ into $x$ in the power series, but I'm having trouble moving on from there.






real-analysis







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  • 1




    $begingroup$
    What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
    $endgroup$
    – egreg
    Dec 11 '18 at 23:35
















0












$begingroup$


We consider the power series $sum_{n=1}^{infty} a_nx^n$ where $a_n = frac{2^k}{k}$ if $n$ is even, and $0$ otherwise.



I understand how to find that this series has radius of convergence $R = frac{1}{sqrt{2}}$. What I don't understand is how to now find its interval of convergence. I know I'm supposed to plug $R$ into $x$ in the power series, but I'm having trouble moving on from there.






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
    $endgroup$
    – egreg
    Dec 11 '18 at 23:35














0












0








0





$begingroup$


We consider the power series $sum_{n=1}^{infty} a_nx^n$ where $a_n = frac{2^k}{k}$ if $n$ is even, and $0$ otherwise.



I understand how to find that this series has radius of convergence $R = frac{1}{sqrt{2}}$. What I don't understand is how to now find its interval of convergence. I know I'm supposed to plug $R$ into $x$ in the power series, but I'm having trouble moving on from there.






real-analysis







share|cite|improve this question









$endgroup$




We consider the power series $sum_{n=1}^{infty} a_nx^n$ where $a_n = frac{2^k}{k}$ if $n$ is even, and $0$ otherwise.



I understand how to find that this series has radius of convergence $R = frac{1}{sqrt{2}}$. What I don't understand is how to now find its interval of convergence. I know I'm supposed to plug $R$ into $x$ in the power series, but I'm having trouble moving on from there.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 23:17









JakeJake

646




646








  • 1




    $begingroup$
    What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
    $endgroup$
    – egreg
    Dec 11 '18 at 23:35














  • 1




    $begingroup$
    What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
    $endgroup$
    – egreg
    Dec 11 '18 at 23:35








1




1




$begingroup$
What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
$endgroup$
– egreg
Dec 11 '18 at 23:35




$begingroup$
What's $k$? Is it perhaps $k=n/2$ when $n$ is even?
$endgroup$
– egreg
Dec 11 '18 at 23:35










3 Answers
3






active

oldest

votes


















1












$begingroup$

I assume that
$$
a_n=begin{cases}
dfrac{2^k}{k} & text{if $n=2k$ is even} \[4px]
0 & text{if $n$ is odd}
end{cases}
$$
Then you can rewrite your series as
$$
f(x)=sum_{k=1}^{infty} frac{2^k}{k}x^{2k}
$$
Apply the ratio test (for $xne0$):
$$
left|frac{2^{k+1}x^{2k+2}/(k+1)}{2^kx^{2k}/k}right|=frac{2(k+1)}{k}|x^2|
$$
which has limit $2|x^2|=2x^2$. We know that the series converges when the limit is $<1$ and diverges when the limit is $>1$, so the series converges for
$$
2x^2<1
$$
that is, $-1/sqrt{2}<x<1/sqrt{2}$. The radius of convergence is thus $1/sqrt{2}$.



Hadamard's criterion says that the radius of convergence is
$$
limsup_{ntoinfty}sqrt[n]{a_n}
$$
which is usually more complicated to compute. But when the ratio test or the root test is successful, the radius of convergence can be determined more easily.



If you consider
$$
g(y)=sum_{k=1}^{infty}frac{2^k}{k}y^k
$$
which has radius of convergence $1/2$, then, for $|y|<1/2$,
$$
g'(y)=sum_{k=1}^infty 2^ky^{k-1}=2sum_{k=0}^infty (2y)^k=frac{2}{1-2y}
$$
(geometric series).



It follows that
$$
g(y)=-log(1-2y)+c
$$
Since $g(0)=0$, we conclude $g(y)=-log(1-2y)$. Therefore your series can be summed as well, because $f(x)=g(x^2)$, yielding
$$
f(x)=-log(1-2x^2)
$$
The series converges only on $(-1/sqrt{2},1/sqrt{2})$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The interval of convergence is $(c-R, c+R)$, where $c$ is the center of the series and $R$ is the radius of convergence. The series may also converge at $R$ or $-R$, which you have to check independently. Your series is centered at $0$.



    Edit: Why? A power series will always converge at $c$ (the sum will just be $a_0$). Now if the series converges elsewhere, then $exists r in (0,infty)$ s.t. the series converges whenever $|x-c|<r$ and diverges otherwise. This is stated here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
      $endgroup$
      – Jake
      Dec 11 '18 at 23:21












    • $begingroup$
      @Jake see my edit
      $endgroup$
      – Dando18
      Dec 11 '18 at 23:28



















    0












    $begingroup$

    By what Frenchmen call Abel’s lemma (I do not know about other countries), the interval of convergence is the set of all $x$ such that the power series converges. It is contained in $[-R,R]$ and contains $(-R,R)$.
    So you just have to check whether the series converges for $x=R$ and $x=-R$.



    Edit: On the other hand, you might want to check your computations, because, as is, $R=1/2$.






    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









      1












      $begingroup$

      I assume that
      $$
      a_n=begin{cases}
      dfrac{2^k}{k} & text{if $n=2k$ is even} \[4px]
      0 & text{if $n$ is odd}
      end{cases}
      $$
      Then you can rewrite your series as
      $$
      f(x)=sum_{k=1}^{infty} frac{2^k}{k}x^{2k}
      $$
      Apply the ratio test (for $xne0$):
      $$
      left|frac{2^{k+1}x^{2k+2}/(k+1)}{2^kx^{2k}/k}right|=frac{2(k+1)}{k}|x^2|
      $$
      which has limit $2|x^2|=2x^2$. We know that the series converges when the limit is $<1$ and diverges when the limit is $>1$, so the series converges for
      $$
      2x^2<1
      $$
      that is, $-1/sqrt{2}<x<1/sqrt{2}$. The radius of convergence is thus $1/sqrt{2}$.



      Hadamard's criterion says that the radius of convergence is
      $$
      limsup_{ntoinfty}sqrt[n]{a_n}
      $$
      which is usually more complicated to compute. But when the ratio test or the root test is successful, the radius of convergence can be determined more easily.



      If you consider
      $$
      g(y)=sum_{k=1}^{infty}frac{2^k}{k}y^k
      $$
      which has radius of convergence $1/2$, then, for $|y|<1/2$,
      $$
      g'(y)=sum_{k=1}^infty 2^ky^{k-1}=2sum_{k=0}^infty (2y)^k=frac{2}{1-2y}
      $$
      (geometric series).



      It follows that
      $$
      g(y)=-log(1-2y)+c
      $$
      Since $g(0)=0$, we conclude $g(y)=-log(1-2y)$. Therefore your series can be summed as well, because $f(x)=g(x^2)$, yielding
      $$
      f(x)=-log(1-2x^2)
      $$
      The series converges only on $(-1/sqrt{2},1/sqrt{2})$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I assume that
        $$
        a_n=begin{cases}
        dfrac{2^k}{k} & text{if $n=2k$ is even} \[4px]
        0 & text{if $n$ is odd}
        end{cases}
        $$
        Then you can rewrite your series as
        $$
        f(x)=sum_{k=1}^{infty} frac{2^k}{k}x^{2k}
        $$
        Apply the ratio test (for $xne0$):
        $$
        left|frac{2^{k+1}x^{2k+2}/(k+1)}{2^kx^{2k}/k}right|=frac{2(k+1)}{k}|x^2|
        $$
        which has limit $2|x^2|=2x^2$. We know that the series converges when the limit is $<1$ and diverges when the limit is $>1$, so the series converges for
        $$
        2x^2<1
        $$
        that is, $-1/sqrt{2}<x<1/sqrt{2}$. The radius of convergence is thus $1/sqrt{2}$.



        Hadamard's criterion says that the radius of convergence is
        $$
        limsup_{ntoinfty}sqrt[n]{a_n}
        $$
        which is usually more complicated to compute. But when the ratio test or the root test is successful, the radius of convergence can be determined more easily.



        If you consider
        $$
        g(y)=sum_{k=1}^{infty}frac{2^k}{k}y^k
        $$
        which has radius of convergence $1/2$, then, for $|y|<1/2$,
        $$
        g'(y)=sum_{k=1}^infty 2^ky^{k-1}=2sum_{k=0}^infty (2y)^k=frac{2}{1-2y}
        $$
        (geometric series).



        It follows that
        $$
        g(y)=-log(1-2y)+c
        $$
        Since $g(0)=0$, we conclude $g(y)=-log(1-2y)$. Therefore your series can be summed as well, because $f(x)=g(x^2)$, yielding
        $$
        f(x)=-log(1-2x^2)
        $$
        The series converges only on $(-1/sqrt{2},1/sqrt{2})$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I assume that
          $$
          a_n=begin{cases}
          dfrac{2^k}{k} & text{if $n=2k$ is even} \[4px]
          0 & text{if $n$ is odd}
          end{cases}
          $$
          Then you can rewrite your series as
          $$
          f(x)=sum_{k=1}^{infty} frac{2^k}{k}x^{2k}
          $$
          Apply the ratio test (for $xne0$):
          $$
          left|frac{2^{k+1}x^{2k+2}/(k+1)}{2^kx^{2k}/k}right|=frac{2(k+1)}{k}|x^2|
          $$
          which has limit $2|x^2|=2x^2$. We know that the series converges when the limit is $<1$ and diverges when the limit is $>1$, so the series converges for
          $$
          2x^2<1
          $$
          that is, $-1/sqrt{2}<x<1/sqrt{2}$. The radius of convergence is thus $1/sqrt{2}$.



          Hadamard's criterion says that the radius of convergence is
          $$
          limsup_{ntoinfty}sqrt[n]{a_n}
          $$
          which is usually more complicated to compute. But when the ratio test or the root test is successful, the radius of convergence can be determined more easily.



          If you consider
          $$
          g(y)=sum_{k=1}^{infty}frac{2^k}{k}y^k
          $$
          which has radius of convergence $1/2$, then, for $|y|<1/2$,
          $$
          g'(y)=sum_{k=1}^infty 2^ky^{k-1}=2sum_{k=0}^infty (2y)^k=frac{2}{1-2y}
          $$
          (geometric series).



          It follows that
          $$
          g(y)=-log(1-2y)+c
          $$
          Since $g(0)=0$, we conclude $g(y)=-log(1-2y)$. Therefore your series can be summed as well, because $f(x)=g(x^2)$, yielding
          $$
          f(x)=-log(1-2x^2)
          $$
          The series converges only on $(-1/sqrt{2},1/sqrt{2})$.






          share|cite|improve this answer









          $endgroup$



          I assume that
          $$
          a_n=begin{cases}
          dfrac{2^k}{k} & text{if $n=2k$ is even} \[4px]
          0 & text{if $n$ is odd}
          end{cases}
          $$
          Then you can rewrite your series as
          $$
          f(x)=sum_{k=1}^{infty} frac{2^k}{k}x^{2k}
          $$
          Apply the ratio test (for $xne0$):
          $$
          left|frac{2^{k+1}x^{2k+2}/(k+1)}{2^kx^{2k}/k}right|=frac{2(k+1)}{k}|x^2|
          $$
          which has limit $2|x^2|=2x^2$. We know that the series converges when the limit is $<1$ and diverges when the limit is $>1$, so the series converges for
          $$
          2x^2<1
          $$
          that is, $-1/sqrt{2}<x<1/sqrt{2}$. The radius of convergence is thus $1/sqrt{2}$.



          Hadamard's criterion says that the radius of convergence is
          $$
          limsup_{ntoinfty}sqrt[n]{a_n}
          $$
          which is usually more complicated to compute. But when the ratio test or the root test is successful, the radius of convergence can be determined more easily.



          If you consider
          $$
          g(y)=sum_{k=1}^{infty}frac{2^k}{k}y^k
          $$
          which has radius of convergence $1/2$, then, for $|y|<1/2$,
          $$
          g'(y)=sum_{k=1}^infty 2^ky^{k-1}=2sum_{k=0}^infty (2y)^k=frac{2}{1-2y}
          $$
          (geometric series).



          It follows that
          $$
          g(y)=-log(1-2y)+c
          $$
          Since $g(0)=0$, we conclude $g(y)=-log(1-2y)$. Therefore your series can be summed as well, because $f(x)=g(x^2)$, yielding
          $$
          f(x)=-log(1-2x^2)
          $$
          The series converges only on $(-1/sqrt{2},1/sqrt{2})$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 23:51









          egregegreg

          181k1485203




          181k1485203























              0












              $begingroup$

              The interval of convergence is $(c-R, c+R)$, where $c$ is the center of the series and $R$ is the radius of convergence. The series may also converge at $R$ or $-R$, which you have to check independently. Your series is centered at $0$.



              Edit: Why? A power series will always converge at $c$ (the sum will just be $a_0$). Now if the series converges elsewhere, then $exists r in (0,infty)$ s.t. the series converges whenever $|x-c|<r$ and diverges otherwise. This is stated here.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
                $endgroup$
                – Jake
                Dec 11 '18 at 23:21












              • $begingroup$
                @Jake see my edit
                $endgroup$
                – Dando18
                Dec 11 '18 at 23:28
















              0












              $begingroup$

              The interval of convergence is $(c-R, c+R)$, where $c$ is the center of the series and $R$ is the radius of convergence. The series may also converge at $R$ or $-R$, which you have to check independently. Your series is centered at $0$.



              Edit: Why? A power series will always converge at $c$ (the sum will just be $a_0$). Now if the series converges elsewhere, then $exists r in (0,infty)$ s.t. the series converges whenever $|x-c|<r$ and diverges otherwise. This is stated here.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
                $endgroup$
                – Jake
                Dec 11 '18 at 23:21












              • $begingroup$
                @Jake see my edit
                $endgroup$
                – Dando18
                Dec 11 '18 at 23:28














              0












              0








              0





              $begingroup$

              The interval of convergence is $(c-R, c+R)$, where $c$ is the center of the series and $R$ is the radius of convergence. The series may also converge at $R$ or $-R$, which you have to check independently. Your series is centered at $0$.



              Edit: Why? A power series will always converge at $c$ (the sum will just be $a_0$). Now if the series converges elsewhere, then $exists r in (0,infty)$ s.t. the series converges whenever $|x-c|<r$ and diverges otherwise. This is stated here.






              share|cite|improve this answer











              $endgroup$



              The interval of convergence is $(c-R, c+R)$, where $c$ is the center of the series and $R$ is the radius of convergence. The series may also converge at $R$ or $-R$, which you have to check independently. Your series is centered at $0$.



              Edit: Why? A power series will always converge at $c$ (the sum will just be $a_0$). Now if the series converges elsewhere, then $exists r in (0,infty)$ s.t. the series converges whenever $|x-c|<r$ and diverges otherwise. This is stated here.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 11 '18 at 23:27

























              answered Dec 11 '18 at 23:19









              Dando18Dando18

              4,67741235




              4,67741235












              • $begingroup$
                In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
                $endgroup$
                – Jake
                Dec 11 '18 at 23:21












              • $begingroup$
                @Jake see my edit
                $endgroup$
                – Dando18
                Dec 11 '18 at 23:28


















              • $begingroup$
                In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
                $endgroup$
                – Jake
                Dec 11 '18 at 23:21












              • $begingroup$
                @Jake see my edit
                $endgroup$
                – Dando18
                Dec 11 '18 at 23:28
















              $begingroup$
              In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
              $endgroup$
              – Jake
              Dec 11 '18 at 23:21






              $begingroup$
              In this case it's $(c - R, c + R)$, though, since the power series diverges for $x = pm frac{1}{sqrt{2}}$. I'm having trouble understanding why this is true.
              $endgroup$
              – Jake
              Dec 11 '18 at 23:21














              $begingroup$
              @Jake see my edit
              $endgroup$
              – Dando18
              Dec 11 '18 at 23:28




              $begingroup$
              @Jake see my edit
              $endgroup$
              – Dando18
              Dec 11 '18 at 23:28











              0












              $begingroup$

              By what Frenchmen call Abel’s lemma (I do not know about other countries), the interval of convergence is the set of all $x$ such that the power series converges. It is contained in $[-R,R]$ and contains $(-R,R)$.
              So you just have to check whether the series converges for $x=R$ and $x=-R$.



              Edit: On the other hand, you might want to check your computations, because, as is, $R=1/2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                By what Frenchmen call Abel’s lemma (I do not know about other countries), the interval of convergence is the set of all $x$ such that the power series converges. It is contained in $[-R,R]$ and contains $(-R,R)$.
                So you just have to check whether the series converges for $x=R$ and $x=-R$.



                Edit: On the other hand, you might want to check your computations, because, as is, $R=1/2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  By what Frenchmen call Abel’s lemma (I do not know about other countries), the interval of convergence is the set of all $x$ such that the power series converges. It is contained in $[-R,R]$ and contains $(-R,R)$.
                  So you just have to check whether the series converges for $x=R$ and $x=-R$.



                  Edit: On the other hand, you might want to check your computations, because, as is, $R=1/2$.






                  share|cite|improve this answer









                  $endgroup$



                  By what Frenchmen call Abel’s lemma (I do not know about other countries), the interval of convergence is the set of all $x$ such that the power series converges. It is contained in $[-R,R]$ and contains $(-R,R)$.
                  So you just have to check whether the series converges for $x=R$ and $x=-R$.



                  Edit: On the other hand, you might want to check your computations, because, as is, $R=1/2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 23:27









                  MindlackMindlack

                  3,53717




                  3,53717






























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                      Tonle Sap (See)

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                      Tonle Sap Satellitenfoto (April 1994) Geographische Lage Kambodscha Zuflüsse Tonle-Sap-Fluss, Siem Reap River Abfluss Tonle-Sap-Fluss Orte in der Nähe Siem Reap, Battambang Daten Koordinaten 12° 52′ 34″  N , 104° 4′ 23″  O 12.876 104.073 Koordinaten: 12° 52′ 34″  N , 104° 4′ 23″  O Höhe über Meeresspiegel f1 0,5 Meter Fläche 2.700–25.000 km² dep1 f5 Länge 250 km (maximal) dep1 f6 Breite 100 km (maximal) dep1 f7 Volumen 80 km³ (maximal) dep1 f8 Maximale Tiefe 14 m f10 Mittlere Tiefe 2–3 m (minimal) f11 Besonderheiten größter See Südostasiens Der Tonle-Sap-See (Khmer: បឹង​ទន្លេសាប , Aussprache: [ boeung tunleː saːp ]) in Kambodscha ist der größte See Südostasiens und eines der fischreichsten Binnengewässer der Erde. Inhaltsverzeichnis 1 Geographie 2 Wirtschaftliche Bedeutung 2.1 Verkehr 2.2 Reis 2.3 Fisch 3 Ökologie und Gefährdung 3.1 Überfischung...
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                      I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

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                      0 <?php $servername = "localhost"; $username="root"; $password=""; $dbName="escape_room"; $user_username = $_POST['Input_user']; $user_password = $_POST['Input_pass']; $conn = new mysqli($servername, $username, $password, $dbName); if(!$conn){ die("Cound not Connect: " . mysqli_connect_error()); } $sql = "SELECT pass FROM escape_room WHERE user = '".$user_username."' "; $result = mysqli_query($conn, $sql); //Get the result and confirm login if(mysqli_num_rows($result)>0){ while($row = mysqli_fetch_assoc($result)){ if($row['pass'] == $user_password){ echo "login success"; ...
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                      Guatemaltekische Davis-Cup-Mannschaft

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                      Guatemala Kapitän Anthony Vásquez Aktuelles ITF-Ranking 73 Statistik Erste Teilnahme 1990 Davis-Cup-Teilnahmen 22 Bestes Ergebnis Amerika-Gruppenzone II HF (1995, 2000, 2001) Ewige Bilanz 42:30 Erfolgreichste Spieler Meiste Siege gesamt Daniel Chávez Morales (29) Meiste Einzelsiege Cristian Paiz (21) Meiste Doppelsiege Daniel Chávez Morales (17) Bestes Doppel Jacobo Chávez / Daniel Chávez Morales (9) Meiste Teilnahmen Cristian Paiz (32) Meiste Jahre Daniel Chávez Morales (10) Letzte Aktualisierung der Infobox: 3. Juni 2012 Die guatemaltekische Davis-Cup-Mannschaft ist die Tennisnationalmannschaft Guatemalas. Geschichte | 1990 nahm Guatemala erstmals am Davis Cup teil. Das beste Resultat erreichte die Mannschaft jeweils 1995, 2000 und 2001 mit dem Halbfinale in der Amerika-Gruppenzone II. Bester Spieler ist Daniel Chávez Morales mit 29 Siegen bei insgesamt 31 Teilnahmen. Rekord...
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