Convergence of $sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$

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Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$



As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$



How to handle $(-n)^lflooralpharfloor$ with that result?






real-analysis sequences-and-series convergence







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  • i think you should take two cases, for $lflooralpharfloor$ to be odd or even
    – Subhajit Halder
    Sep 6 at 12:49














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Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$



As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$



How to handle $(-n)^lflooralpharfloor$ with that result?






real-analysis sequences-and-series convergence







share|cite|improve this question























  • i think you should take two cases, for $lflooralpharfloor$ to be odd or even
    – Subhajit Halder
    Sep 6 at 12:49












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$



As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$



How to handle $(-n)^lflooralpharfloor$ with that result?






real-analysis sequences-and-series convergence







share|cite|improve this question















Convergence of the following series as $alphainmathbbR$
$$sum_n=1^infty(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1$$



As $n to +infty$ we have that $fracsin^3left(frac1nright)left(log(n)right)^alpha-1 = mathcalOleft(frac1n^3log^alpha-1(n)right)$



How to handle $(-n)^lflooralpharfloor$ with that result?






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






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edited Sep 6 at 13:27









gimusi

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asked Sep 6 at 11:52









F.inc

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  • i think you should take two cases, for $lflooralpharfloor$ to be odd or even
    – Subhajit Halder
    Sep 6 at 12:49
















  • i think you should take two cases, for $lflooralpharfloor$ to be odd or even
    – Subhajit Halder
    Sep 6 at 12:49















i think you should take two cases, for $lflooralpharfloor$ to be odd or even
– Subhajit Halder
Sep 6 at 12:49




i think you should take two cases, for $lflooralpharfloor$ to be odd or even
– Subhajit Halder
Sep 6 at 12:49










1 Answer
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Yes the key point is that



$$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$



then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then



$$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$



which converges by limit comparison test with $sum frac1n^2$.



And you can proceed similarly for the others values.






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    up vote
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    Yes the key point is that



    $$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$



    then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then



    $$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$



    which converges by limit comparison test with $sum frac1n^2$.



    And you can proceed similarly for the others values.






    share|cite|improve this answer


























      up vote
      2
      down vote













      Yes the key point is that



      $$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$



      then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then



      $$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$



      which converges by limit comparison test with $sum frac1n^2$.



      And you can proceed similarly for the others values.






      share|cite|improve this answer
























        up vote
        2
        down vote










        up vote
        2
        down vote









        Yes the key point is that



        $$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$



        then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then



        $$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$



        which converges by limit comparison test with $sum frac1n^2$.



        And you can proceed similarly for the others values.






        share|cite|improve this answer














        Yes the key point is that



        $$fracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim frac1n^3log^alpha-1n$$



        then we need to distinguish the cases, for example for $0le alpha <1$ we have $lflooralpha rfloor=0$ and then



        $$(-n)^lflooralpharfloorfracsin^3left(frac1nright)left(log(n)right)^alpha-1 sim fraclog nn^3$$



        which converges by limit comparison test with $sum frac1n^2$.



        And you can proceed similarly for the others values.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 6 at 14:08

























        answered Sep 6 at 13:23









        gimusi

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