Series convergence: $sin (n frac{pi}{2})$












2












$begingroup$



Determine whether the following series :
$$sum_{n=1}^infty sin left(frac{npi}{2}right) frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.




I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the value of $sin$ will go from $1$ to $-1$.



Could I theoretically reduce this series into a subseries:



$$sum_{n=0}^infty sin left(frac{(2n+1)pi}{2}right) frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$



And then treat it as if it were a standard alternating series?






calculus sequences-and-series convergence







share|cite|improve this question











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  • $begingroup$
    You can reduce the series. Show using Cauchy Criterion.
    $endgroup$
    – Melody
    Dec 3 '18 at 17:26
















2












$begingroup$



Determine whether the following series :
$$sum_{n=1}^infty sin left(frac{npi}{2}right) frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.




I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the value of $sin$ will go from $1$ to $-1$.



Could I theoretically reduce this series into a subseries:



$$sum_{n=0}^infty sin left(frac{(2n+1)pi}{2}right) frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$



And then treat it as if it were a standard alternating series?






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can reduce the series. Show using Cauchy Criterion.
    $endgroup$
    – Melody
    Dec 3 '18 at 17:26














2












2








2





$begingroup$



Determine whether the following series :
$$sum_{n=1}^infty sin left(frac{npi}{2}right) frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.




I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the value of $sin$ will go from $1$ to $-1$.



Could I theoretically reduce this series into a subseries:



$$sum_{n=0}^infty sin left(frac{(2n+1)pi}{2}right) frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$



And then treat it as if it were a standard alternating series?






calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$





Determine whether the following series :
$$sum_{n=1}^infty sin left(frac{npi}{2}right) frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.




I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the value of $sin$ will go from $1$ to $-1$.



Could I theoretically reduce this series into a subseries:



$$sum_{n=0}^infty sin left(frac{(2n+1)pi}{2}right) frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$



And then treat it as if it were a standard alternating series?






calculus sequences-and-series convergence




calculus sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 6:35









choco_addicted

8,06261847




8,06261847










asked Dec 3 '18 at 17:21









J. LastinJ. Lastin

965




965












  • $begingroup$
    You can reduce the series. Show using Cauchy Criterion.
    $endgroup$
    – Melody
    Dec 3 '18 at 17:26


















  • $begingroup$
    You can reduce the series. Show using Cauchy Criterion.
    $endgroup$
    – Melody
    Dec 3 '18 at 17:26
















$begingroup$
You can reduce the series. Show using Cauchy Criterion.
$endgroup$
– Melody
Dec 3 '18 at 17:26




$begingroup$
You can reduce the series. Show using Cauchy Criterion.
$endgroup$
– Melody
Dec 3 '18 at 17:26










2 Answers
2






active

oldest

votes


















1












$begingroup$

HINT




  • Note that
    $$
    frac{n^2+2}{n^3+n}
    = frac{1}{n} times frac{n^2+2}{n^2+1}
    = frac{1}{n} left[ 1 + frac{1}{n^2+1} right]
    = Theta(1/n)
    $$
    Does this series converge?

  • Adding $sin(npi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The alternating series should converge right?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:05






  • 1




    $begingroup$
    @J.Lastin indeed
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:07










  • $begingroup$
    Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:09








  • 1




    $begingroup$
    @J.Lastin also correct
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:56



















2












$begingroup$

Guide:




  • We have $$sum_{n=1}^infty sin (frac{npi}{2}) frac{n^2+2}{n^3 +n}= sum_{n=1}^infty(-1)^{n+1} frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$


  • Try alternating series test.







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









    1












    $begingroup$

    HINT




    • Note that
      $$
      frac{n^2+2}{n^3+n}
      = frac{1}{n} times frac{n^2+2}{n^2+1}
      = frac{1}{n} left[ 1 + frac{1}{n^2+1} right]
      = Theta(1/n)
      $$
      Does this series converge?

    • Adding $sin(npi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The alternating series should converge right?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:05






    • 1




      $begingroup$
      @J.Lastin indeed
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:07










    • $begingroup$
      Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:09








    • 1




      $begingroup$
      @J.Lastin also correct
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:56
















    1












    $begingroup$

    HINT




    • Note that
      $$
      frac{n^2+2}{n^3+n}
      = frac{1}{n} times frac{n^2+2}{n^2+1}
      = frac{1}{n} left[ 1 + frac{1}{n^2+1} right]
      = Theta(1/n)
      $$
      Does this series converge?

    • Adding $sin(npi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The alternating series should converge right?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:05






    • 1




      $begingroup$
      @J.Lastin indeed
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:07










    • $begingroup$
      Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:09








    • 1




      $begingroup$
      @J.Lastin also correct
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:56














    1












    1








    1





    $begingroup$

    HINT




    • Note that
      $$
      frac{n^2+2}{n^3+n}
      = frac{1}{n} times frac{n^2+2}{n^2+1}
      = frac{1}{n} left[ 1 + frac{1}{n^2+1} right]
      = Theta(1/n)
      $$
      Does this series converge?

    • Adding $sin(npi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?






    share|cite|improve this answer









    $endgroup$



    HINT




    • Note that
      $$
      frac{n^2+2}{n^3+n}
      = frac{1}{n} times frac{n^2+2}{n^2+1}
      = frac{1}{n} left[ 1 + frac{1}{n^2+1} right]
      = Theta(1/n)
      $$
      Does this series converge?

    • Adding $sin(npi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 17:27









    gt6989bgt6989b

    33.7k22455




    33.7k22455












    • $begingroup$
      The alternating series should converge right?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:05






    • 1




      $begingroup$
      @J.Lastin indeed
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:07










    • $begingroup$
      Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:09








    • 1




      $begingroup$
      @J.Lastin also correct
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:56


















    • $begingroup$
      The alternating series should converge right?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:05






    • 1




      $begingroup$
      @J.Lastin indeed
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:07










    • $begingroup$
      Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
      $endgroup$
      – J. Lastin
      Dec 3 '18 at 18:09








    • 1




      $begingroup$
      @J.Lastin also correct
      $endgroup$
      – gt6989b
      Dec 3 '18 at 18:56
















    $begingroup$
    The alternating series should converge right?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:05




    $begingroup$
    The alternating series should converge right?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:05




    1




    1




    $begingroup$
    @J.Lastin indeed
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:07




    $begingroup$
    @J.Lastin indeed
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:07












    $begingroup$
    Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:09






    $begingroup$
    Okay and if i want to check for absolute convergence, can I do limit comparison with $frac{1}{n}$, meaning that the series converges only conditionally?
    $endgroup$
    – J. Lastin
    Dec 3 '18 at 18:09






    1




    1




    $begingroup$
    @J.Lastin also correct
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:56




    $begingroup$
    @J.Lastin also correct
    $endgroup$
    – gt6989b
    Dec 3 '18 at 18:56











    2












    $begingroup$

    Guide:




    • We have $$sum_{n=1}^infty sin (frac{npi}{2}) frac{n^2+2}{n^3 +n}= sum_{n=1}^infty(-1)^{n+1} frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$


    • Try alternating series test.







    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Guide:




      • We have $$sum_{n=1}^infty sin (frac{npi}{2}) frac{n^2+2}{n^3 +n}= sum_{n=1}^infty(-1)^{n+1} frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$


      • Try alternating series test.







      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Guide:




        • We have $$sum_{n=1}^infty sin (frac{npi}{2}) frac{n^2+2}{n^3 +n}= sum_{n=1}^infty(-1)^{n+1} frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$


        • Try alternating series test.







        share|cite|improve this answer









        $endgroup$



        Guide:




        • We have $$sum_{n=1}^infty sin (frac{npi}{2}) frac{n^2+2}{n^3 +n}= sum_{n=1}^infty(-1)^{n+1} frac{(2n-1)^2+2}{(2n-1)^3 +(2n-1)}$$


        • Try alternating series test.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 17:28









        Siong Thye GohSiong Thye Goh

        100k1465117




        100k1465117






























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