Study the convergence of the series $sum_{n=1}^infty left( sqrt[17]{5+ frac1n} - sqrt[17]{5}right)^{a}q^n$...












1












$begingroup$


I have a problem with this task, because I think the most important is idea to do convergence of the series
$$
sum_{n=1}^inftyleft( sqrt[17]{5+ frac{1}{n} } - sqrt[17]{5}right)^{!a},
$$
but it is difficult for me because it is power $a$ and for $a=1$.



I can use claim about three series, but in this case I completely don't knew what can I do.






real-analysis calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have a problem with this task, because I think the most important is idea to do convergence of the series
    $$
    sum_{n=1}^inftyleft( sqrt[17]{5+ frac{1}{n} } - sqrt[17]{5}right)^{!a},
    $$
    but it is difficult for me because it is power $a$ and for $a=1$.



    I can use claim about three series, but in this case I completely don't knew what can I do.






    real-analysis calculus sequences-and-series convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have a problem with this task, because I think the most important is idea to do convergence of the series
      $$
      sum_{n=1}^inftyleft( sqrt[17]{5+ frac{1}{n} } - sqrt[17]{5}right)^{!a},
      $$
      but it is difficult for me because it is power $a$ and for $a=1$.



      I can use claim about three series, but in this case I completely don't knew what can I do.






      real-analysis calculus sequences-and-series convergence







      share|cite|improve this question











      $endgroup$




      I have a problem with this task, because I think the most important is idea to do convergence of the series
      $$
      sum_{n=1}^inftyleft( sqrt[17]{5+ frac{1}{n} } - sqrt[17]{5}right)^{!a},
      $$
      but it is difficult for me because it is power $a$ and for $a=1$.



      I can use claim about three series, but in this case I completely don't knew what can I do.






      real-analysis calculus sequences-and-series convergence




      real-analysis 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 Jan 4 at 17:18









      Yiorgos S. Smyrlis

      63.6k1385165




      63.6k1385165










      asked Jan 1 at 20:29









      MP3129MP3129

      641110




      641110






















          1 Answer
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          active

          oldest

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          5












          $begingroup$

          Hint. First show that
          $$
          lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85},
          $$
          and hence
          $$
          left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)^aapprox left(frac{sqrt[17]{5}}{85}right)^an^{-a}
          $$
          Thus, we have convergence of the series if $|q|<1$, or $q=1$ and $a>1$, or $q=-1$ and $a>0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
            $endgroup$
            – MP3129
            Jan 1 at 22:59








          • 1




            $begingroup$
            If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
            $endgroup$
            – JV.Stalker
            Jan 2 at 4:46










          • $begingroup$
            Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
            $endgroup$
            – MP3129
            Jan 2 at 10:32














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          5












          $begingroup$

          Hint. First show that
          $$
          lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85},
          $$
          and hence
          $$
          left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)^aapprox left(frac{sqrt[17]{5}}{85}right)^an^{-a}
          $$
          Thus, we have convergence of the series if $|q|<1$, or $q=1$ and $a>1$, or $q=-1$ and $a>0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
            $endgroup$
            – MP3129
            Jan 1 at 22:59








          • 1




            $begingroup$
            If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
            $endgroup$
            – JV.Stalker
            Jan 2 at 4:46










          • $begingroup$
            Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
            $endgroup$
            – MP3129
            Jan 2 at 10:32


















          5












          $begingroup$

          Hint. First show that
          $$
          lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85},
          $$
          and hence
          $$
          left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)^aapprox left(frac{sqrt[17]{5}}{85}right)^an^{-a}
          $$
          Thus, we have convergence of the series if $|q|<1$, or $q=1$ and $a>1$, or $q=-1$ and $a>0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
            $endgroup$
            – MP3129
            Jan 1 at 22:59








          • 1




            $begingroup$
            If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
            $endgroup$
            – JV.Stalker
            Jan 2 at 4:46










          • $begingroup$
            Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
            $endgroup$
            – MP3129
            Jan 2 at 10:32
















          5












          5








          5





          $begingroup$

          Hint. First show that
          $$
          lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85},
          $$
          and hence
          $$
          left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)^aapprox left(frac{sqrt[17]{5}}{85}right)^an^{-a}
          $$
          Thus, we have convergence of the series if $|q|<1$, or $q=1$ and $a>1$, or $q=-1$ and $a>0$.






          share|cite|improve this answer









          $endgroup$



          Hint. First show that
          $$
          lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85},
          $$
          and hence
          $$
          left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)^aapprox left(frac{sqrt[17]{5}}{85}right)^an^{-a}
          $$
          Thus, we have convergence of the series if $|q|<1$, or $q=1$ and $a>1$, or $q=-1$ and $a>0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 1 at 20:37









          Yiorgos S. SmyrlisYiorgos S. Smyrlis

          63.6k1385165




          63.6k1385165












          • $begingroup$
            Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
            $endgroup$
            – MP3129
            Jan 1 at 22:59








          • 1




            $begingroup$
            If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
            $endgroup$
            – JV.Stalker
            Jan 2 at 4:46










          • $begingroup$
            Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
            $endgroup$
            – MP3129
            Jan 2 at 10:32




















          • $begingroup$
            Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
            $endgroup$
            – MP3129
            Jan 1 at 22:59








          • 1




            $begingroup$
            If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
            $endgroup$
            – JV.Stalker
            Jan 2 at 4:46










          • $begingroup$
            Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
            $endgroup$
            – MP3129
            Jan 2 at 10:32


















          $begingroup$
          Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
          $endgroup$
          – MP3129
          Jan 1 at 22:59






          $begingroup$
          Ok, but if I have to show that $lim_{ntoinfty,}nleft(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=frac{sqrt[17]{5}}{85}$ I should multiply by coupling $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)$ but I think it is impossible for power $1/17$ - I can do it for example for $1/16$ but not for $1/17$
          $endgroup$
          – MP3129
          Jan 1 at 22:59






          1




          1




          $begingroup$
          If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
          $endgroup$
          – JV.Stalker
          Jan 2 at 4:46




          $begingroup$
          If I understand well, you need the following additional information: Pulling out $sqrt[17]{5}$ $left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5}right)=left(sqrt[17]{5}sqrt[17]{1+frac{1}{5n}}-sqrt[17]{5}right)=sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)$ Using Bernoulli's inequality you receive an upper bound: $sqrt[17]{5}left(sqrt[17]{1+frac{1}{5n}}-1right)lesqrt[17]{5}left({1+frac{1}{85n}}-1right)=frac{sqrt[17]{5}}{85n}$
          $endgroup$
          – JV.Stalker
          Jan 2 at 4:46












          $begingroup$
          Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
          $endgroup$
          – MP3129
          Jan 2 at 10:32






          $begingroup$
          Ok, thanks, but I need also lower limit, but I think the only possibility is $ge left(sqrt[17]{5+frac{1}{n}}-sqrt[17]{5+frac{1}{n^{b}}}right)$ for $b>1$, but in this case I cannot show that the limit of this expression is $frac{sqrt[17]{5}}{85n}$ because here I also need 3 ranks so I should find an expression where the limit is obvious
          $endgroup$
          – MP3129
          Jan 2 at 10:32




















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