By Cauchy I theorem on limits of sequence I am getting sequence is divergent












0












$begingroup$


By Cauchy II Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



Then, $Lim_{ntoinfty}$$(S_{1}S_{2}$...$S_{n})^{frac{1}{n}}$=l



And , By Cauchy I Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



$Lim_{ntoinfty}$${frac{(S_{1}+S_{2}+...+S_{n})}{n}}$=l



Now, for $Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$ is convergent by using Cauchy II theorem ,as



$a_{n}=big(frac{1}{2}big)^{n} implies lim_{ntoinfty}a_{n}=big(frac{1}{2}big)^{n}=0$



But is not convergent by Cauchy I Theorem ,



Let $t= Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$



taking logarithm on both sides ,we get



$log(t)={frac{1}{n}}Big[logbig(frac{1}{2}big)+logbig(frac{1}{2}big)^{2}+...+logbig(frac{1}{2}big)^{n}Big]$



By using Cauchy I Theorem ,we get



$a_{n}=logbig(frac{1}{2}big)^{n}=n * logbig(frac{1}{2}big)$



$Lim_{ntoinfty}a_{n}=Lim_{ntoinfty}n * logbig(frac{1}{2}big)=infty $



$implies lim_{nto infty}log(t)=infty$



$t=e^{infty}=infty$



$Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}=infty$ .



Hence sequence diverges !!



By Cauchy I theorem on limits of sequence I am getting sequence is divergent and but using second I am getting it is convergent .



Please help me tell me where I am getting wrong !!






real-analysis sequences-and-series







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  • $begingroup$
    Please type out the part relevant to the question using MathJax and do away with the image.
    $endgroup$
    – Swapnil Rustagi
    Jan 25 at 13:51










  • $begingroup$
    Please tell me, where i have done gone wrong .
    $endgroup$
    – sejy
    Jan 25 at 16:03
















0












$begingroup$


By Cauchy II Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



Then, $Lim_{ntoinfty}$$(S_{1}S_{2}$...$S_{n})^{frac{1}{n}}$=l



And , By Cauchy I Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



$Lim_{ntoinfty}$${frac{(S_{1}+S_{2}+...+S_{n})}{n}}$=l



Now, for $Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$ is convergent by using Cauchy II theorem ,as



$a_{n}=big(frac{1}{2}big)^{n} implies lim_{ntoinfty}a_{n}=big(frac{1}{2}big)^{n}=0$



But is not convergent by Cauchy I Theorem ,



Let $t= Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$



taking logarithm on both sides ,we get



$log(t)={frac{1}{n}}Big[logbig(frac{1}{2}big)+logbig(frac{1}{2}big)^{2}+...+logbig(frac{1}{2}big)^{n}Big]$



By using Cauchy I Theorem ,we get



$a_{n}=logbig(frac{1}{2}big)^{n}=n * logbig(frac{1}{2}big)$



$Lim_{ntoinfty}a_{n}=Lim_{ntoinfty}n * logbig(frac{1}{2}big)=infty $



$implies lim_{nto infty}log(t)=infty$



$t=e^{infty}=infty$



$Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}=infty$ .



Hence sequence diverges !!



By Cauchy I theorem on limits of sequence I am getting sequence is divergent and but using second I am getting it is convergent .



Please help me tell me where I am getting wrong !!






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please type out the part relevant to the question using MathJax and do away with the image.
    $endgroup$
    – Swapnil Rustagi
    Jan 25 at 13:51










  • $begingroup$
    Please tell me, where i have done gone wrong .
    $endgroup$
    – sejy
    Jan 25 at 16:03














0












0








0





$begingroup$


By Cauchy II Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



Then, $Lim_{ntoinfty}$$(S_{1}S_{2}$...$S_{n})^{frac{1}{n}}$=l



And , By Cauchy I Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



$Lim_{ntoinfty}$${frac{(S_{1}+S_{2}+...+S_{n})}{n}}$=l



Now, for $Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$ is convergent by using Cauchy II theorem ,as



$a_{n}=big(frac{1}{2}big)^{n} implies lim_{ntoinfty}a_{n}=big(frac{1}{2}big)^{n}=0$



But is not convergent by Cauchy I Theorem ,



Let $t= Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$



taking logarithm on both sides ,we get



$log(t)={frac{1}{n}}Big[logbig(frac{1}{2}big)+logbig(frac{1}{2}big)^{2}+...+logbig(frac{1}{2}big)^{n}Big]$



By using Cauchy I Theorem ,we get



$a_{n}=logbig(frac{1}{2}big)^{n}=n * logbig(frac{1}{2}big)$



$Lim_{ntoinfty}a_{n}=Lim_{ntoinfty}n * logbig(frac{1}{2}big)=infty $



$implies lim_{nto infty}log(t)=infty$



$t=e^{infty}=infty$



$Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}=infty$ .



Hence sequence diverges !!



By Cauchy I theorem on limits of sequence I am getting sequence is divergent and but using second I am getting it is convergent .



Please help me tell me where I am getting wrong !!






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




By Cauchy II Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



Then, $Lim_{ntoinfty}$$(S_{1}S_{2}$...$S_{n})^{frac{1}{n}}$=l



And , By Cauchy I Theorem on Limits ,



If $<S_{n}>$ is a sequence such that $S_{n}>0$ , $forall n$ and $Lim_{ntoinfty}S_{n}=l$



$Lim_{ntoinfty}$${frac{(S_{1}+S_{2}+...+S_{n})}{n}}$=l



Now, for $Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$ is convergent by using Cauchy II theorem ,as



$a_{n}=big(frac{1}{2}big)^{n} implies lim_{ntoinfty}a_{n}=big(frac{1}{2}big)^{n}=0$



But is not convergent by Cauchy I Theorem ,



Let $t= Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}$



taking logarithm on both sides ,we get



$log(t)={frac{1}{n}}Big[logbig(frac{1}{2}big)+logbig(frac{1}{2}big)^{2}+...+logbig(frac{1}{2}big)^{n}Big]$



By using Cauchy I Theorem ,we get



$a_{n}=logbig(frac{1}{2}big)^{n}=n * logbig(frac{1}{2}big)$



$Lim_{ntoinfty}a_{n}=Lim_{ntoinfty}n * logbig(frac{1}{2}big)=infty $



$implies lim_{nto infty}log(t)=infty$



$t=e^{infty}=infty$



$Lim_{ntoinfty}Big[big(frac{1}{2}big) big(frac{1}{2}big)^{2}...big(frac{1}{2}big)^{n}Big]^{frac{1}{n}}=infty$ .



Hence sequence diverges !!



By Cauchy I theorem on limits of sequence I am getting sequence is divergent and but using second I am getting it is convergent .



Please help me tell me where I am getting wrong !!






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 16:02







sejy

















asked Jan 25 at 12:40









sejysejy

1549




1549












  • $begingroup$
    Please type out the part relevant to the question using MathJax and do away with the image.
    $endgroup$
    – Swapnil Rustagi
    Jan 25 at 13:51










  • $begingroup$
    Please tell me, where i have done gone wrong .
    $endgroup$
    – sejy
    Jan 25 at 16:03


















  • $begingroup$
    Please type out the part relevant to the question using MathJax and do away with the image.
    $endgroup$
    – Swapnil Rustagi
    Jan 25 at 13:51










  • $begingroup$
    Please tell me, where i have done gone wrong .
    $endgroup$
    – sejy
    Jan 25 at 16:03
















$begingroup$
Please type out the part relevant to the question using MathJax and do away with the image.
$endgroup$
– Swapnil Rustagi
Jan 25 at 13:51




$begingroup$
Please type out the part relevant to the question using MathJax and do away with the image.
$endgroup$
– Swapnil Rustagi
Jan 25 at 13:51












$begingroup$
Please tell me, where i have done gone wrong .
$endgroup$
– sejy
Jan 25 at 16:03




$begingroup$
Please tell me, where i have done gone wrong .
$endgroup$
– sejy
Jan 25 at 16:03










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$begingroup$

$log(1/2) <0$ so
$nlog(1/2) to -infty$.






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    $begingroup$

    $log(1/2) <0$ so
    $nlog(1/2) to -infty$.






    share|cite|improve this answer









    $endgroup$


















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      $begingroup$

      $log(1/2) <0$ so
      $nlog(1/2) to -infty$.






      share|cite|improve this answer









      $endgroup$
















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        1





        $begingroup$

        $log(1/2) <0$ so
        $nlog(1/2) to -infty$.






        share|cite|improve this answer









        $endgroup$



        $log(1/2) <0$ so
        $nlog(1/2) to -infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 16:13









        marty cohenmarty cohen

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