Convergence test for a series











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Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.




My attempt:



$$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$



Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.










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    Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.




    My attempt:



    $$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$



    Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.




      My attempt:



      $$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$



      Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.










      share|cite|improve this question
















      Let S = $sum_{n=1}^{infty}{frac{n^k}{((n^3+n)^{frac{1}{3}}-n)}},forallspacespacespacespace kinmathbb{Z}.$ Say if this series converge of diverge.




      My attempt:



      $$S=sum_{n=1}^{infty}frac{(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2}{n^{1-k}}$$



      Ratio test in undecisive, and Raabe-Duhmal test gave me again $1$ so what test should I use? It is clear that the general term of these series $to0$ iff $1-k<2implies k>1$ and divergent for $k<1$.







      sequences-and-series real-numbers numerical-calculus






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      edited Nov 14 at 11:23









      Crazy for maths

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      asked Nov 14 at 11:09









      C. Cristi

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          Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.






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            May be, you could start considering that, for large values of $n$,
            $$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$






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              2 Answers
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              Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.






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                Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.






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                  Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.






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                  Note thatbegin{align}frac1{sqrt[3]{n^3+n}-n}&=frac1{sqrt[3]{n^3+n}-sqrt[3]{n^3}}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}{n^3+n-n^3}\&=frac{sqrt[3]{(n^3+n)^2}+sqrt[3]{(n^3+n)n^3}+sqrt[3]{(n^3)^2}}n\&inleft[3n,frac{3sqrt[3]{(n^3+n)^2}}nright].end{align}This is enough to prove that the series converges if and only if $k<-2$.







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                  edited Nov 14 at 11:30

























                  answered Nov 14 at 11:19









                  José Carlos Santos

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                      May be, you could start considering that, for large values of $n$,
                      $$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$






                      share|cite|improve this answer



























                        up vote
                        0
                        down vote













                        May be, you could start considering that, for large values of $n$,
                        $$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          May be, you could start considering that, for large values of $n$,
                          $$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$






                          share|cite|improve this answer














                          May be, you could start considering that, for large values of $n$,
                          $$(n^3+n)^{frac{2}{3}}+n(n^3+n)^{frac{1}{3}}+n^2=3 n^2+1-frac{2}{9 n^2}+Oleft(frac{1}{n^4}right)$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 14 at 11:45

























                          answered Nov 14 at 11:35









                          Claude Leibovici

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