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Question about convergence of infinite products


Infinite products (involving complex numbers)Convergence of a productHelp with Convergence/DivergenceA basic sequence convergence questionQuestion about series increasing and convergenceProof about the convergence of a sequenceStatements regarding sequences and seriesInfinite series: Integral test( Proof )Question regarding implication of series convergence.Convergence of the ratio of two infinite series.













-1












$begingroup$


Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!










share|cite|improve this question











$endgroup$











  • $begingroup$
    What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
    $endgroup$
    – Yves Daoust
    Mar 14 at 16:52











  • $begingroup$
    $prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
    $endgroup$
    – manifolded
    Mar 14 at 17:06










  • $begingroup$
    "bounded above" makes it a very different question !
    $endgroup$
    – Yves Daoust
    Mar 14 at 17:10










  • $begingroup$
    True, my bad, don't know why I did that. Thanks for the answer.
    $endgroup$
    – manifolded
    Mar 14 at 17:11















-1












$begingroup$


Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!










share|cite|improve this question











$endgroup$











  • $begingroup$
    What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
    $endgroup$
    – Yves Daoust
    Mar 14 at 16:52











  • $begingroup$
    $prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
    $endgroup$
    – manifolded
    Mar 14 at 17:06










  • $begingroup$
    "bounded above" makes it a very different question !
    $endgroup$
    – Yves Daoust
    Mar 14 at 17:10










  • $begingroup$
    True, my bad, don't know why I did that. Thanks for the answer.
    $endgroup$
    – manifolded
    Mar 14 at 17:11













-1












-1








-1





$begingroup$


Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!










share|cite|improve this question











$endgroup$




Because $lim_nrightarrow infty(1+fracxn)^n = e^x$, is it true that $prod_n(1+x_n)$ is bounded above by $e^sum_n=1^infty x_n$ and hence $prod_n(1+x_n)$ converges or diverges depending on $sum_n=1^inftyx_n$? Thanks!







calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 17:08







manifolded

















asked Mar 14 at 16:18









manifoldedmanifolded

48819




48819











  • $begingroup$
    What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
    $endgroup$
    – Yves Daoust
    Mar 14 at 16:52











  • $begingroup$
    $prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
    $endgroup$
    – manifolded
    Mar 14 at 17:06










  • $begingroup$
    "bounded above" makes it a very different question !
    $endgroup$
    – Yves Daoust
    Mar 14 at 17:10










  • $begingroup$
    True, my bad, don't know why I did that. Thanks for the answer.
    $endgroup$
    – manifolded
    Mar 14 at 17:11
















  • $begingroup$
    What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
    $endgroup$
    – Yves Daoust
    Mar 14 at 16:52











  • $begingroup$
    $prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
    $endgroup$
    – manifolded
    Mar 14 at 17:06










  • $begingroup$
    "bounded above" makes it a very different question !
    $endgroup$
    – Yves Daoust
    Mar 14 at 17:10










  • $begingroup$
    True, my bad, don't know why I did that. Thanks for the answer.
    $endgroup$
    – manifolded
    Mar 14 at 17:11















$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52





$begingroup$
What do you mean by $$lim_ntoinftyprod_nleft(1+x_nright)$$ exactly ?
$endgroup$
– Yves Daoust
Mar 14 at 16:52













$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06




$begingroup$
$prod_n (1+x_n) = (1+x_1)(1+x_2)..$ and by the limit I mean, is this product bounded above by $e^sum x_n$
$endgroup$
– manifolded
Mar 14 at 17:06












$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10




$begingroup$
"bounded above" makes it a very different question !
$endgroup$
– Yves Daoust
Mar 14 at 17:10












$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11




$begingroup$
True, my bad, don't know why I did that. Thanks for the answer.
$endgroup$
– manifolded
Mar 14 at 17:11










1 Answer
1






active

oldest

votes


















1












$begingroup$

If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$



and there is no reason that



$$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$




Given the change of the question, this answer is irrelevant.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$



    and there is no reason that



    $$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$




    Given the change of the question, this answer is irrelevant.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$



      and there is no reason that



      $$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$




      Given the change of the question, this answer is irrelevant.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$



        and there is no reason that



        $$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$




        Given the change of the question, this answer is irrelevant.






        share|cite|improve this answer









        $endgroup$



        If $|x_n|<1$, $$sum_n=1^mlog(1+x_n)=sum_n=1^mx_n-frac12sum_n=1^mx_n^2+frac13sum_n=1^mx_n^3-cdots$$



        and there is no reason that



        $$lim_mtoinftysum_n=1^mlog(1+x_n)=lim_mtoinftysum_n=1^mx_n.$$




        Given the change of the question, this answer is irrelevant.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 at 17:09









        Yves DaoustYves Daoust

        131k676229




        131k676229



























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