$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$ convergence/divergence












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I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.






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  • $begingroup$
    @Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
    $endgroup$
    – Andarrkor
    Dec 24 '18 at 20:57


















0












$begingroup$


I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
    $endgroup$
    – Andarrkor
    Dec 24 '18 at 20:57
















0












0








0





$begingroup$


I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.






sequences-and-series







share|cite|improve this question









$endgroup$




I am trying to analyze the convergence/divergence of:
$sum_{n=0}^infty (sqrt[3]{n+1} - sqrt[3]{n})^alpha (2+n^4)^beta$
depending on $alpha$ and $beta$. This is how far I have gone:
First i have changed it a bit using $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ and i have studied this: $sum_{n=0}^infty frac{(2+n^4)^beta}{((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}$ . I used the limit-convergence test
$lim_{nto infty} frac{(2+n^4)^beta}{n^lambda((n+1)^{2/3} + (n(n+1))^{1/3} + n^{2/3})^alpha}=l$ and now I would have to chose $lambda$ so that $0lt l lt infty$. My problem comes now. If i choose $lambda= 4beta-frac{2}{3}alpha$ the limit shoud be finite and bigger than $0$. So now, would it be correct to say that if $-1 gt 4beta-frac{2}{3}alpha$ the series is convergent? If it is not correct, how should the problem be done? Thanks in advance.






sequences-and-series




sequences-and-series






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asked Dec 24 '18 at 20:20









AndarrkorAndarrkor

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  • $begingroup$
    @Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
    $endgroup$
    – Andarrkor
    Dec 24 '18 at 20:57




















  • $begingroup$
    @Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
    $endgroup$
    – Andarrkor
    Dec 24 '18 at 20:57


















$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57






$begingroup$
@Clayton $a=sqrt[3]{n+1}$ and $b=sqrt[3]{n}$
$endgroup$
– Andarrkor
Dec 24 '18 at 20:57












1 Answer
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$begingroup$

It's much simpler using equivalents





  • $sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
    $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$


  • $(2+n^4)^beta sim_infty n^{4beta}.$
    Therefore,
    $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
    and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.






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    $begingroup$

    It's much simpler using equivalents





    • $sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
      $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$


    • $(2+n^4)^beta sim_infty n^{4beta}.$
      Therefore,
      $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
      and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It's much simpler using equivalents





      • $sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
        $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$


      • $(2+n^4)^beta sim_infty n^{4beta}.$
        Therefore,
        $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
        and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It's much simpler using equivalents





        • $sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
          $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$


        • $(2+n^4)^beta sim_infty n^{4beta}.$
          Therefore,
          $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
          and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.






        share|cite|improve this answer









        $endgroup$



        It's much simpler using equivalents





        • $sqrt[3]{n+1}-sqrt[3]{n}=sqrt[3]{n}Bigl(sqrt[3]{1+dfrac 1n}-1Bigr)=sqrt[3]{n}dfrac1{3n}=dfrac1{3n^{tfrac23}}$, so
          $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alphasim_{infty}dfrac1{3^alpha,n^{tfrac{2alpha}3}}.$$


        • $(2+n^4)^beta sim_infty n^{4beta}.$
          Therefore,
          $$bigl(sqrt[3]{n+1} - sqrt[3]{n}bigr)^alpha (2+n^4)^betasim_infty dfrac1{3^alpha,n^{tfrac{2alpha}3-4beta}}$$
          and the latter, which is a $p$-series, converges if and only if $;dfrac{2alpha}3-4beta>1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 21:26









        BernardBernard

        121k740116




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