Finding values of x for which series converges











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Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.



For which values of $x in mathbb{R}$ does this series converge?



I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.






real-analysis sequences-and-series







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    up vote
    0
    down vote

    favorite












    Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.



    For which values of $x in mathbb{R}$ does this series converge?



    I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.






    real-analysis sequences-and-series







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.



      For which values of $x in mathbb{R}$ does this series converge?



      I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.






      real-analysis sequences-and-series







      share|cite|improve this question













      Consider $a_n = (-1)^n frac{x^2+n}{n^2}$ for $n in mathbb{N}$.



      For which values of $x in mathbb{R}$ does this series converge?



      I applies ratio test and get 1, which does not imply anything clearly. I am not sure if root test does any help or not.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 18 at 14:48









      ChakSayantan

      986




      986






















          2 Answers
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          This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.



          $$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$



          Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.






          share|cite|improve this answer




























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            The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$






            share|cite|improve this answer























            • "all the conditions" meaning $a_nto 0$ as $nto infty$
              – Mason
              Nov 18 at 15:26










            • The absolute value of term is decreasing and converges to zero.
              – Mohammad Riazi-Kermani
              Nov 18 at 15:40











            Your Answer





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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            up vote
            0
            down vote



            accepted










            This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.



            $$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$



            Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.



              $$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$



              Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.



                $$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$



                Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.






                share|cite|improve this answer












                This converges for all values of $x$. And we compute exactly what it converges to through some algebraic manipulation.



                $$sum_{n=1}^infty(-1)^nfrac{x^2+n}{n^2}=x^2sum_{n=1}^infty frac{ (-1)^n}{n^2}+ sum_{n=1}^inftyfrac{(-1)^n}{n}=-x^2eta(2)-eta(1)=frac{-pi^2}{12}x^2-ln(2)$$



                Where $eta(s)=sum_{n=1}^infty (-1)^{n+1}/n^s$ which is the dirichlet eta function. Here's a graph.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 18 at 15:48









                Mason

                1,8251425




                1,8251425






















                    up vote
                    0
                    down vote













                    The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$






                    share|cite|improve this answer























                    • "all the conditions" meaning $a_nto 0$ as $nto infty$
                      – Mason
                      Nov 18 at 15:26










                    • The absolute value of term is decreasing and converges to zero.
                      – Mohammad Riazi-Kermani
                      Nov 18 at 15:40















                    up vote
                    0
                    down vote













                    The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$






                    share|cite|improve this answer























                    • "all the conditions" meaning $a_nto 0$ as $nto infty$
                      – Mason
                      Nov 18 at 15:26










                    • The absolute value of term is decreasing and converges to zero.
                      – Mohammad Riazi-Kermani
                      Nov 18 at 15:40













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$






                    share|cite|improve this answer














                    The series $$sum _{n=1} ^infty a_n= sum _{n=1} ^infty(−1)^nfrac { x^2+n}{n^2} $$ satisfies all the conditions of a convergent alternating series thus it converges for all values of $xin R$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 18 at 15:36









                    Mason

                    1,8251425




                    1,8251425










                    answered Nov 18 at 14:59









                    Mohammad Riazi-Kermani

                    40.3k41958




                    40.3k41958












                    • "all the conditions" meaning $a_nto 0$ as $nto infty$
                      – Mason
                      Nov 18 at 15:26










                    • The absolute value of term is decreasing and converges to zero.
                      – Mohammad Riazi-Kermani
                      Nov 18 at 15:40


















                    • "all the conditions" meaning $a_nto 0$ as $nto infty$
                      – Mason
                      Nov 18 at 15:26










                    • The absolute value of term is decreasing and converges to zero.
                      – Mohammad Riazi-Kermani
                      Nov 18 at 15:40
















                    "all the conditions" meaning $a_nto 0$ as $nto infty$
                    – Mason
                    Nov 18 at 15:26




                    "all the conditions" meaning $a_nto 0$ as $nto infty$
                    – Mason
                    Nov 18 at 15:26












                    The absolute value of term is decreasing and converges to zero.
                    – Mohammad Riazi-Kermani
                    Nov 18 at 15:40




                    The absolute value of term is decreasing and converges to zero.
                    – Mohammad Riazi-Kermani
                    Nov 18 at 15:40


















                     

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