Find all the $c ge 0$ for which $sum_{n=1}^{+ infty }a _{n}$ is absolutely convergent












2












$begingroup$


We consider: $$a_{1}=c-1$$ $$a_{n+1}= frac{-n}{n+c cdot sqrt[n]{ln(n^{9876}+17)}}cdot a_{n}, nge 1$$ $$cge0$$
I want to use Rabbe Test, because then in a simple way it comes out that the series is convergent for every $c>1$. However I have two doubts:
1) Raabe Test is for $a_{n}>0$, but if I do $r_{n}=n(|frac{a_{n}}{a_{n+1}}|-1)$ I knew that I must removing minus at $n$ and then I can leave the module. Hovewer I'm not sure if it's allowed. 2) If Raabe Test is a good way to do this task I knew only when my series is convergent, but I don't knew when is absolutely convergent so the more I do not know if Raabe Test is a good idea.






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    $begingroup$


    We consider: $$a_{1}=c-1$$ $$a_{n+1}= frac{-n}{n+c cdot sqrt[n]{ln(n^{9876}+17)}}cdot a_{n}, nge 1$$ $$cge0$$
    I want to use Rabbe Test, because then in a simple way it comes out that the series is convergent for every $c>1$. However I have two doubts:
    1) Raabe Test is for $a_{n}>0$, but if I do $r_{n}=n(|frac{a_{n}}{a_{n+1}}|-1)$ I knew that I must removing minus at $n$ and then I can leave the module. Hovewer I'm not sure if it's allowed. 2) If Raabe Test is a good way to do this task I knew only when my series is convergent, but I don't knew when is absolutely convergent so the more I do not know if Raabe Test is a good idea.






    real-analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      We consider: $$a_{1}=c-1$$ $$a_{n+1}= frac{-n}{n+c cdot sqrt[n]{ln(n^{9876}+17)}}cdot a_{n}, nge 1$$ $$cge0$$
      I want to use Rabbe Test, because then in a simple way it comes out that the series is convergent for every $c>1$. However I have two doubts:
      1) Raabe Test is for $a_{n}>0$, but if I do $r_{n}=n(|frac{a_{n}}{a_{n+1}}|-1)$ I knew that I must removing minus at $n$ and then I can leave the module. Hovewer I'm not sure if it's allowed. 2) If Raabe Test is a good way to do this task I knew only when my series is convergent, but I don't knew when is absolutely convergent so the more I do not know if Raabe Test is a good idea.






      real-analysis







      share|cite|improve this question











      $endgroup$




      We consider: $$a_{1}=c-1$$ $$a_{n+1}= frac{-n}{n+c cdot sqrt[n]{ln(n^{9876}+17)}}cdot a_{n}, nge 1$$ $$cge0$$
      I want to use Rabbe Test, because then in a simple way it comes out that the series is convergent for every $c>1$. However I have two doubts:
      1) Raabe Test is for $a_{n}>0$, but if I do $r_{n}=n(|frac{a_{n}}{a_{n+1}}|-1)$ I knew that I must removing minus at $n$ and then I can leave the module. Hovewer I'm not sure if it's allowed. 2) If Raabe Test is a good way to do this task I knew only when my series is convergent, but I don't knew when is absolutely convergent so the more I do not know if Raabe Test is a good idea.






      real-analysis




      real-analysis






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 16:53







      MP3129

















      asked Dec 31 '18 at 16:36









      MP3129MP3129

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      36010






















          1 Answer
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          $begingroup$

          Define $b_n := | a_n |$, and observe that $sum_{n geq 1} a_n$ is absolutely convergent iff $sum_{n geq 1} b_n$ is convergent.



          If $c neq 1$, then the $b_n$'s are positive, so we may use Raabe's test to test the convergence of $sum_{n geq 1} b_n$. This will tell us about whether or not $sum_{n geq 1} a_n$ is absolutely convergent.



          Carrying this out, we have
          $$ n left( frac{b_n}{b_{n+1}} - 1right) = c sqrt[n]{log n^{9876} + 17} to c {rm as } n to infty.$$



          So $sum_{n geq 1} a_n$ is absolutely convergent if $c > 1$ and not absolutely convergent if $c < 1$.



          The $c = 1$ case is special. When $c = 1$, we have $ a_n = 0$ for all $n geq 1$. So the series is trivially absolutely convergent - it sums to zero.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
            $endgroup$
            – Jakobian
            Dec 31 '18 at 17:17












          • $begingroup$
            @Jakobian Sorry, I reorganised the answer.
            $endgroup$
            – Kenny Wong
            Dec 31 '18 at 17:32











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Define $b_n := | a_n |$, and observe that $sum_{n geq 1} a_n$ is absolutely convergent iff $sum_{n geq 1} b_n$ is convergent.



          If $c neq 1$, then the $b_n$'s are positive, so we may use Raabe's test to test the convergence of $sum_{n geq 1} b_n$. This will tell us about whether or not $sum_{n geq 1} a_n$ is absolutely convergent.



          Carrying this out, we have
          $$ n left( frac{b_n}{b_{n+1}} - 1right) = c sqrt[n]{log n^{9876} + 17} to c {rm as } n to infty.$$



          So $sum_{n geq 1} a_n$ is absolutely convergent if $c > 1$ and not absolutely convergent if $c < 1$.



          The $c = 1$ case is special. When $c = 1$, we have $ a_n = 0$ for all $n geq 1$. So the series is trivially absolutely convergent - it sums to zero.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
            $endgroup$
            – Jakobian
            Dec 31 '18 at 17:17












          • $begingroup$
            @Jakobian Sorry, I reorganised the answer.
            $endgroup$
            – Kenny Wong
            Dec 31 '18 at 17:32
















          2












          $begingroup$

          Define $b_n := | a_n |$, and observe that $sum_{n geq 1} a_n$ is absolutely convergent iff $sum_{n geq 1} b_n$ is convergent.



          If $c neq 1$, then the $b_n$'s are positive, so we may use Raabe's test to test the convergence of $sum_{n geq 1} b_n$. This will tell us about whether or not $sum_{n geq 1} a_n$ is absolutely convergent.



          Carrying this out, we have
          $$ n left( frac{b_n}{b_{n+1}} - 1right) = c sqrt[n]{log n^{9876} + 17} to c {rm as } n to infty.$$



          So $sum_{n geq 1} a_n$ is absolutely convergent if $c > 1$ and not absolutely convergent if $c < 1$.



          The $c = 1$ case is special. When $c = 1$, we have $ a_n = 0$ for all $n geq 1$. So the series is trivially absolutely convergent - it sums to zero.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
            $endgroup$
            – Jakobian
            Dec 31 '18 at 17:17












          • $begingroup$
            @Jakobian Sorry, I reorganised the answer.
            $endgroup$
            – Kenny Wong
            Dec 31 '18 at 17:32














          2












          2








          2





          $begingroup$

          Define $b_n := | a_n |$, and observe that $sum_{n geq 1} a_n$ is absolutely convergent iff $sum_{n geq 1} b_n$ is convergent.



          If $c neq 1$, then the $b_n$'s are positive, so we may use Raabe's test to test the convergence of $sum_{n geq 1} b_n$. This will tell us about whether or not $sum_{n geq 1} a_n$ is absolutely convergent.



          Carrying this out, we have
          $$ n left( frac{b_n}{b_{n+1}} - 1right) = c sqrt[n]{log n^{9876} + 17} to c {rm as } n to infty.$$



          So $sum_{n geq 1} a_n$ is absolutely convergent if $c > 1$ and not absolutely convergent if $c < 1$.



          The $c = 1$ case is special. When $c = 1$, we have $ a_n = 0$ for all $n geq 1$. So the series is trivially absolutely convergent - it sums to zero.






          share|cite|improve this answer











          $endgroup$



          Define $b_n := | a_n |$, and observe that $sum_{n geq 1} a_n$ is absolutely convergent iff $sum_{n geq 1} b_n$ is convergent.



          If $c neq 1$, then the $b_n$'s are positive, so we may use Raabe's test to test the convergence of $sum_{n geq 1} b_n$. This will tell us about whether or not $sum_{n geq 1} a_n$ is absolutely convergent.



          Carrying this out, we have
          $$ n left( frac{b_n}{b_{n+1}} - 1right) = c sqrt[n]{log n^{9876} + 17} to c {rm as } n to infty.$$



          So $sum_{n geq 1} a_n$ is absolutely convergent if $c > 1$ and not absolutely convergent if $c < 1$.



          The $c = 1$ case is special. When $c = 1$, we have $ a_n = 0$ for all $n geq 1$. So the series is trivially absolutely convergent - it sums to zero.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 31 '18 at 17:36









          Jakobian

          2,650721




          2,650721










          answered Dec 31 '18 at 16:42









          Kenny WongKenny Wong

          19k21440




          19k21440












          • $begingroup$
            $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
            $endgroup$
            – Jakobian
            Dec 31 '18 at 17:17












          • $begingroup$
            @Jakobian Sorry, I reorganised the answer.
            $endgroup$
            – Kenny Wong
            Dec 31 '18 at 17:32


















          • $begingroup$
            $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
            $endgroup$
            – Jakobian
            Dec 31 '18 at 17:17












          • $begingroup$
            @Jakobian Sorry, I reorganised the answer.
            $endgroup$
            – Kenny Wong
            Dec 31 '18 at 17:32
















          $begingroup$
          $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
          $endgroup$
          – Jakobian
          Dec 31 '18 at 17:17






          $begingroup$
          $b_n$'s are positive when $cneq 1$. When $c=1$ we can't use Raabe's test
          $endgroup$
          – Jakobian
          Dec 31 '18 at 17:17














          $begingroup$
          @Jakobian Sorry, I reorganised the answer.
          $endgroup$
          – Kenny Wong
          Dec 31 '18 at 17:32




          $begingroup$
          @Jakobian Sorry, I reorganised the answer.
          $endgroup$
          – Kenny Wong
          Dec 31 '18 at 17:32


















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