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How can I know a value of this limit?


Limits in integrationEvaluation of a limitCan this limit be solved algebraically?How can i resolve this limit without L'Hopital's Rule?find this limit without l'hopital's ruleHow to find the value of $a$ and $b$ from this limit problem with or without L'Hopital's formula?How to do limit?Find the limit of the sequence $(4^n)/((2n)!)$How to solve this limit involving sine and log?Find a limit, no Taylor formula













1












$begingroup$


$$
lim_tto 1^- (1-t) sum_n=0^infty fract^n1+t^n
$$



I try to use L'hopital's rule but not successfully. Also I know that exist formula Sonine but I can't understand how I can use it.






real-analysis limits







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
    $endgroup$
    – cand
    yesterday







  • 1




    $begingroup$
    also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
    $endgroup$
    – cand
    yesterday










  • $begingroup$
    took closest interpretation, please check
    $endgroup$
    – gt6989b
    yesterday










  • $begingroup$
    it is lim t→1−0
    $endgroup$
    – Andrey Komisarov
    yesterday






  • 1




    $begingroup$
    @cand: the sum diverges for $tgt1$.
    $endgroup$
    – robjohn♦
    5 hours ago















1












$begingroup$


$$
lim_tto 1^- (1-t) sum_n=0^infty fract^n1+t^n
$$



I try to use L'hopital's rule but not successfully. Also I know that exist formula Sonine but I can't understand how I can use it.






real-analysis limits







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
    $endgroup$
    – cand
    yesterday







  • 1




    $begingroup$
    also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
    $endgroup$
    – cand
    yesterday










  • $begingroup$
    took closest interpretation, please check
    $endgroup$
    – gt6989b
    yesterday










  • $begingroup$
    it is lim t→1−0
    $endgroup$
    – Andrey Komisarov
    yesterday






  • 1




    $begingroup$
    @cand: the sum diverges for $tgt1$.
    $endgroup$
    – robjohn♦
    5 hours ago













1












1








1


1



$begingroup$


$$
lim_tto 1^- (1-t) sum_n=0^infty fract^n1+t^n
$$



I try to use L'hopital's rule but not successfully. Also I know that exist formula Sonine but I can't understand how I can use it.






real-analysis limits







share|cite|improve this question











$endgroup$




$$
lim_tto 1^- (1-t) sum_n=0^infty fract^n1+t^n
$$



I try to use L'hopital's rule but not successfully. Also I know that exist formula Sonine but I can't understand how I can use it.






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









robjohn♦

269k27309635




269k27309635










asked yesterday









Andrey KomisarovAndrey Komisarov

353




353







  • 2




    $begingroup$
    is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
    $endgroup$
    – cand
    yesterday







  • 1




    $begingroup$
    also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
    $endgroup$
    – cand
    yesterday










  • $begingroup$
    took closest interpretation, please check
    $endgroup$
    – gt6989b
    yesterday










  • $begingroup$
    it is lim t→1−0
    $endgroup$
    – Andrey Komisarov
    yesterday






  • 1




    $begingroup$
    @cand: the sum diverges for $tgt1$.
    $endgroup$
    – robjohn♦
    5 hours ago












  • 2




    $begingroup$
    is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
    $endgroup$
    – cand
    yesterday







  • 1




    $begingroup$
    also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
    $endgroup$
    – cand
    yesterday










  • $begingroup$
    took closest interpretation, please check
    $endgroup$
    – gt6989b
    yesterday










  • $begingroup$
    it is lim t→1−0
    $endgroup$
    – Andrey Komisarov
    yesterday






  • 1




    $begingroup$
    @cand: the sum diverges for $tgt1$.
    $endgroup$
    – robjohn♦
    5 hours ago







2




2




$begingroup$
is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
$endgroup$
– cand
yesterday





$begingroup$
is it $lim_tto 1^-$ on the limit or simply $lim_tto 1$?
$endgroup$
– cand
yesterday





1




1




$begingroup$
also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
$endgroup$
– cand
yesterday




$begingroup$
also try reading meta.math.stackexchange.com/questions/5020/… to learn how mathjax work so peoples can understand you equation part better.
$endgroup$
– cand
yesterday












$begingroup$
took closest interpretation, please check
$endgroup$
– gt6989b
yesterday




$begingroup$
took closest interpretation, please check
$endgroup$
– gt6989b
yesterday












$begingroup$
it is lim t→1−0
$endgroup$
– Andrey Komisarov
yesterday




$begingroup$
it is lim t→1−0
$endgroup$
– Andrey Komisarov
yesterday




1




1




$begingroup$
@cand: the sum diverges for $tgt1$.
$endgroup$
– robjohn♦
5 hours ago




$begingroup$
@cand: the sum diverges for $tgt1$.
$endgroup$
– robjohn♦
5 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

Fix $tin(0,1)$. Let $f(x)=fract^x1+t^x$. Using area principle, we have that
$$sum_n=0^infty f(n)=int_0^infty f(x) dx+E,$$
where $|E|le f(0)=frac12$.



It remains to estimate the integral. Let $y=t^x$. Then the integral can be rewritten as
$$int_0^infty f(x) dx=int_0^infty fract^x1+t^x dx=frac-1log tint_0^1fracdy1+y=frac-log2log t.$$
So the limit we are looking for is
$$lim_trightarrow1^-(1-t)sum_n=0^infty f(n)=lim_trightarrow1^-fract-1log tlog2=log2.$$






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    For any $m$,
    $$
    beginalign
    (1-t)sum_n=0^inftyfract^n1+t^n
    &=(1-t)sum_n=0^inftysum_k=1^m-1(-1)^k-1t^kn\
    &+(-1)^m-1(1-t)sum_n=0^inftyfract^mn1+t^n\
    &=sum_k=1^m-1(-1)^k-1frac1-t1-t^k+O!left(frac1-t1-t^mright)tag1
    endalign
    $$
    Let $tto1^-$,
    $$
    lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
    =sum_k=1^m-1frac(-1)^k-1k+O!left(frac1mright)tag2
    $$
    Let $mtoinfty$,
    $$
    beginalign
    lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
    &=sum_k=1^inftyfrac(-1)^k-1k\[6pt]
    &=log(2)tag3
    endalign
    $$






    share|cite|improve this answer











    $endgroup$












      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Fix $tin(0,1)$. Let $f(x)=fract^x1+t^x$. Using area principle, we have that
      $$sum_n=0^infty f(n)=int_0^infty f(x) dx+E,$$
      where $|E|le f(0)=frac12$.



      It remains to estimate the integral. Let $y=t^x$. Then the integral can be rewritten as
      $$int_0^infty f(x) dx=int_0^infty fract^x1+t^x dx=frac-1log tint_0^1fracdy1+y=frac-log2log t.$$
      So the limit we are looking for is
      $$lim_trightarrow1^-(1-t)sum_n=0^infty f(n)=lim_trightarrow1^-fract-1log tlog2=log2.$$






      share|cite|improve this answer









      $endgroup$

















        5












        $begingroup$

        Fix $tin(0,1)$. Let $f(x)=fract^x1+t^x$. Using area principle, we have that
        $$sum_n=0^infty f(n)=int_0^infty f(x) dx+E,$$
        where $|E|le f(0)=frac12$.



        It remains to estimate the integral. Let $y=t^x$. Then the integral can be rewritten as
        $$int_0^infty f(x) dx=int_0^infty fract^x1+t^x dx=frac-1log tint_0^1fracdy1+y=frac-log2log t.$$
        So the limit we are looking for is
        $$lim_trightarrow1^-(1-t)sum_n=0^infty f(n)=lim_trightarrow1^-fract-1log tlog2=log2.$$






        share|cite|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          Fix $tin(0,1)$. Let $f(x)=fract^x1+t^x$. Using area principle, we have that
          $$sum_n=0^infty f(n)=int_0^infty f(x) dx+E,$$
          where $|E|le f(0)=frac12$.



          It remains to estimate the integral. Let $y=t^x$. Then the integral can be rewritten as
          $$int_0^infty f(x) dx=int_0^infty fract^x1+t^x dx=frac-1log tint_0^1fracdy1+y=frac-log2log t.$$
          So the limit we are looking for is
          $$lim_trightarrow1^-(1-t)sum_n=0^infty f(n)=lim_trightarrow1^-fract-1log tlog2=log2.$$






          share|cite|improve this answer









          $endgroup$



          Fix $tin(0,1)$. Let $f(x)=fract^x1+t^x$. Using area principle, we have that
          $$sum_n=0^infty f(n)=int_0^infty f(x) dx+E,$$
          where $|E|le f(0)=frac12$.



          It remains to estimate the integral. Let $y=t^x$. Then the integral can be rewritten as
          $$int_0^infty f(x) dx=int_0^infty fract^x1+t^x dx=frac-1log tint_0^1fracdy1+y=frac-log2log t.$$
          So the limit we are looking for is
          $$lim_trightarrow1^-(1-t)sum_n=0^infty f(n)=lim_trightarrow1^-fract-1log tlog2=log2.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 19 hours ago









          Eric YauEric Yau

          48429




          48429





















              3












              $begingroup$

              For any $m$,
              $$
              beginalign
              (1-t)sum_n=0^inftyfract^n1+t^n
              &=(1-t)sum_n=0^inftysum_k=1^m-1(-1)^k-1t^kn\
              &+(-1)^m-1(1-t)sum_n=0^inftyfract^mn1+t^n\
              &=sum_k=1^m-1(-1)^k-1frac1-t1-t^k+O!left(frac1-t1-t^mright)tag1
              endalign
              $$
              Let $tto1^-$,
              $$
              lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
              =sum_k=1^m-1frac(-1)^k-1k+O!left(frac1mright)tag2
              $$
              Let $mtoinfty$,
              $$
              beginalign
              lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
              &=sum_k=1^inftyfrac(-1)^k-1k\[6pt]
              &=log(2)tag3
              endalign
              $$






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                For any $m$,
                $$
                beginalign
                (1-t)sum_n=0^inftyfract^n1+t^n
                &=(1-t)sum_n=0^inftysum_k=1^m-1(-1)^k-1t^kn\
                &+(-1)^m-1(1-t)sum_n=0^inftyfract^mn1+t^n\
                &=sum_k=1^m-1(-1)^k-1frac1-t1-t^k+O!left(frac1-t1-t^mright)tag1
                endalign
                $$
                Let $tto1^-$,
                $$
                lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                =sum_k=1^m-1frac(-1)^k-1k+O!left(frac1mright)tag2
                $$
                Let $mtoinfty$,
                $$
                beginalign
                lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                &=sum_k=1^inftyfrac(-1)^k-1k\[6pt]
                &=log(2)tag3
                endalign
                $$






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  For any $m$,
                  $$
                  beginalign
                  (1-t)sum_n=0^inftyfract^n1+t^n
                  &=(1-t)sum_n=0^inftysum_k=1^m-1(-1)^k-1t^kn\
                  &+(-1)^m-1(1-t)sum_n=0^inftyfract^mn1+t^n\
                  &=sum_k=1^m-1(-1)^k-1frac1-t1-t^k+O!left(frac1-t1-t^mright)tag1
                  endalign
                  $$
                  Let $tto1^-$,
                  $$
                  lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                  =sum_k=1^m-1frac(-1)^k-1k+O!left(frac1mright)tag2
                  $$
                  Let $mtoinfty$,
                  $$
                  beginalign
                  lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                  &=sum_k=1^inftyfrac(-1)^k-1k\[6pt]
                  &=log(2)tag3
                  endalign
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  For any $m$,
                  $$
                  beginalign
                  (1-t)sum_n=0^inftyfract^n1+t^n
                  &=(1-t)sum_n=0^inftysum_k=1^m-1(-1)^k-1t^kn\
                  &+(-1)^m-1(1-t)sum_n=0^inftyfract^mn1+t^n\
                  &=sum_k=1^m-1(-1)^k-1frac1-t1-t^k+O!left(frac1-t1-t^mright)tag1
                  endalign
                  $$
                  Let $tto1^-$,
                  $$
                  lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                  =sum_k=1^m-1frac(-1)^k-1k+O!left(frac1mright)tag2
                  $$
                  Let $mtoinfty$,
                  $$
                  beginalign
                  lim_tto1^-(1-t)sum_n=0^inftyfract^n1+t^n
                  &=sum_k=1^inftyfrac(-1)^k-1k\[6pt]
                  &=log(2)tag3
                  endalign
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 7 hours ago

























                  answered 13 hours ago









                  robjohn♦robjohn

                  269k27309635




                  269k27309635



























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