General formula of $lim_{n to infty} (1+a_n)^{b_n}$ where $b_n to infty$












2












$begingroup$



What is the general formula of $$lim_{n to infty} (1+a_n)^{b_n} quad text{where} quad lim_{n to infty} b_n = infty?$$




For example we have $$lim_{nto infty} left(1 + frac1nright)^n = e$$



Now I want a general formula for $lim_{n to infty} (1+a_n)^{b_n}$. I know if $a_n to A neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n to 0$. Maybe the answer is $e^c$ where $c = lim _{n to infty} b_n a_n$. But even if this is the answer, I want the proof.



It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.






sequences-and-series limits exponential-function exponentiation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think there's a general formula...
    $endgroup$
    – Yanko
    Jan 7 at 21:45










  • $begingroup$
    @yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
    $endgroup$
    – amir na
    Jan 7 at 21:48










  • $begingroup$
    A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
    $endgroup$
    – Yanko
    Jan 7 at 21:49












  • $begingroup$
    @Yanko yes. If it is really convergent to that
    $endgroup$
    – amir na
    Jan 7 at 21:51
















2












$begingroup$



What is the general formula of $$lim_{n to infty} (1+a_n)^{b_n} quad text{where} quad lim_{n to infty} b_n = infty?$$




For example we have $$lim_{nto infty} left(1 + frac1nright)^n = e$$



Now I want a general formula for $lim_{n to infty} (1+a_n)^{b_n}$. I know if $a_n to A neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n to 0$. Maybe the answer is $e^c$ where $c = lim _{n to infty} b_n a_n$. But even if this is the answer, I want the proof.



It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.






sequences-and-series limits exponential-function exponentiation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think there's a general formula...
    $endgroup$
    – Yanko
    Jan 7 at 21:45










  • $begingroup$
    @yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
    $endgroup$
    – amir na
    Jan 7 at 21:48










  • $begingroup$
    A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
    $endgroup$
    – Yanko
    Jan 7 at 21:49












  • $begingroup$
    @Yanko yes. If it is really convergent to that
    $endgroup$
    – amir na
    Jan 7 at 21:51














2












2








2





$begingroup$



What is the general formula of $$lim_{n to infty} (1+a_n)^{b_n} quad text{where} quad lim_{n to infty} b_n = infty?$$




For example we have $$lim_{nto infty} left(1 + frac1nright)^n = e$$



Now I want a general formula for $lim_{n to infty} (1+a_n)^{b_n}$. I know if $a_n to A neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n to 0$. Maybe the answer is $e^c$ where $c = lim _{n to infty} b_n a_n$. But even if this is the answer, I want the proof.



It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.






sequences-and-series limits exponential-function exponentiation







share|cite|improve this question











$endgroup$





What is the general formula of $$lim_{n to infty} (1+a_n)^{b_n} quad text{where} quad lim_{n to infty} b_n = infty?$$




For example we have $$lim_{nto infty} left(1 + frac1nright)^n = e$$



Now I want a general formula for $lim_{n to infty} (1+a_n)^{b_n}$. I know if $a_n to A neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n to 0$. Maybe the answer is $e^c$ where $c = lim _{n to infty} b_n a_n$. But even if this is the answer, I want the proof.



It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.






sequences-and-series limits exponential-function exponentiation




sequences-and-series limits exponential-function exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:44









gt6989b

33.9k22455




33.9k22455










asked Jan 7 at 21:42









amir naamir na

625




625












  • $begingroup$
    I don't think there's a general formula...
    $endgroup$
    – Yanko
    Jan 7 at 21:45










  • $begingroup$
    @yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
    $endgroup$
    – amir na
    Jan 7 at 21:48










  • $begingroup$
    A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
    $endgroup$
    – Yanko
    Jan 7 at 21:49












  • $begingroup$
    @Yanko yes. If it is really convergent to that
    $endgroup$
    – amir na
    Jan 7 at 21:51


















  • $begingroup$
    I don't think there's a general formula...
    $endgroup$
    – Yanko
    Jan 7 at 21:45










  • $begingroup$
    @yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
    $endgroup$
    – amir na
    Jan 7 at 21:48










  • $begingroup$
    A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
    $endgroup$
    – Yanko
    Jan 7 at 21:49












  • $begingroup$
    @Yanko yes. If it is really convergent to that
    $endgroup$
    – amir na
    Jan 7 at 21:51
















$begingroup$
I don't think there's a general formula...
$endgroup$
– Yanko
Jan 7 at 21:45




$begingroup$
I don't think there's a general formula...
$endgroup$
– Yanko
Jan 7 at 21:45












$begingroup$
@yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
$endgroup$
– amir na
Jan 7 at 21:48




$begingroup$
@yanko maybe there is a formula for at least simple sequences like $p(n) / q(n)$ where $p(n)$ and $q(n)$ are polynomials. for example $(1 - n/(n^2 +2n +10))^n$
$endgroup$
– amir na
Jan 7 at 21:48












$begingroup$
A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
$endgroup$
– Yanko
Jan 7 at 21:49






$begingroup$
A proof for the fact that it converges to $e^c$ if $b_na_nrightarrow c$ is what you look for?
$endgroup$
– Yanko
Jan 7 at 21:49














$begingroup$
@Yanko yes. If it is really convergent to that
$endgroup$
– amir na
Jan 7 at 21:51




$begingroup$
@Yanko yes. If it is really convergent to that
$endgroup$
– amir na
Jan 7 at 21:51










3 Answers
3






active

oldest

votes


















2












$begingroup$

Let $c_n := a_n b_n$ and assume that $0lt a_nrightarrow 0$, $b_nrightarrow infty$. Then define:
$$
A_n:={left(1+a_nright)^{b_n}}=left({left(1+a_nright)^{a_n^{-1}}}right)^{c_n}
$$
From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n rightarrow B$, for some $B$, then $A_nrightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
    $endgroup$
    – Yanko
    Jan 7 at 22:08












  • $begingroup$
    Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
    $endgroup$
    – amir na
    Jan 7 at 22:21



















5












$begingroup$

If the limit does exist, we take logs of both sides, getting
$$
lim_{n to infty} (1+a_n)^{b_n}
= lim_{n to infty} expleft(b_n ln (1+a_n)right)
= exp left( lim_{n to infty} b_n ln (1+a_n)right)
= exp left( lim_{n to infty} frac{ln (1+a_n)}{1/b_n}right)
$$
So if $a_n to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let $b_n,a_n$ such that $lim_n a_nb_n=c$. Moreover assume that $a_nrightarrow 0$.



    Then we have



    $$(1+a_n)^{b_n} = (1+a_n)^{frac{c}{a_n}} cdot (1+a_n)^{b_n-frac{c}{a_n}}$$



    We deal with each term alone:



    First, since $(1+frac{1}{n})^nrightarrow e$ we have that $(1+frac{1}{n})^{cn}rightarrow e^c$. As $a_nrightarrow 0$ we can substitute $a_n$ with $frac{1}{n}$.



    It is therefore enough to show that $(1+a_n)^{b_n-frac{c}{a_n}}rightarrow 0$.



    For every $varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<varepsilon$. This implies that $|b_n-frac{c}{a_n}|<frac{varepsilon}{a_n}$.



    Therefore $(1+a_n)^{b_n-frac{c}{a_n}} < (1+a_n)^{frac{varepsilon}{a_n}}$ and the latter convergence to $e^varepsilon$. As $varepsilon$ is arbitrary small, $e^varepsilon$ is arbitrary close to $1$ and this completes the proof.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $c_n := a_n b_n$ and assume that $0lt a_nrightarrow 0$, $b_nrightarrow infty$. Then define:
      $$
      A_n:={left(1+a_nright)^{b_n}}=left({left(1+a_nright)^{a_n^{-1}}}right)^{c_n}
      $$
      From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n rightarrow B$, for some $B$, then $A_nrightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
        $endgroup$
        – Yanko
        Jan 7 at 22:08












      • $begingroup$
        Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
        $endgroup$
        – amir na
        Jan 7 at 22:21
















      2












      $begingroup$

      Let $c_n := a_n b_n$ and assume that $0lt a_nrightarrow 0$, $b_nrightarrow infty$. Then define:
      $$
      A_n:={left(1+a_nright)^{b_n}}=left({left(1+a_nright)^{a_n^{-1}}}right)^{c_n}
      $$
      From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n rightarrow B$, for some $B$, then $A_nrightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
        $endgroup$
        – Yanko
        Jan 7 at 22:08












      • $begingroup$
        Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
        $endgroup$
        – amir na
        Jan 7 at 22:21














      2












      2








      2





      $begingroup$

      Let $c_n := a_n b_n$ and assume that $0lt a_nrightarrow 0$, $b_nrightarrow infty$. Then define:
      $$
      A_n:={left(1+a_nright)^{b_n}}=left({left(1+a_nright)^{a_n^{-1}}}right)^{c_n}
      $$
      From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n rightarrow B$, for some $B$, then $A_nrightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.






      share|cite|improve this answer











      $endgroup$



      Let $c_n := a_n b_n$ and assume that $0lt a_nrightarrow 0$, $b_nrightarrow infty$. Then define:
      $$
      A_n:={left(1+a_nright)^{b_n}}=left({left(1+a_nright)^{a_n^{-1}}}right)^{c_n}
      $$
      From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n rightarrow B$, for some $B$, then $A_nrightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 7 at 22:10

























      answered Jan 7 at 22:06









      SomeStrangeUserSomeStrangeUser

      1,6401025




      1,6401025












      • $begingroup$
        (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
        $endgroup$
        – Yanko
        Jan 7 at 22:08












      • $begingroup$
        Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
        $endgroup$
        – amir na
        Jan 7 at 22:21


















      • $begingroup$
        (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
        $endgroup$
        – Yanko
        Jan 7 at 22:08












      • $begingroup$
        Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
        $endgroup$
        – amir na
        Jan 7 at 22:21
















      $begingroup$
      (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
      $endgroup$
      – Yanko
      Jan 7 at 22:08






      $begingroup$
      (+1) and this is an overkill. It basically gives kind of a formula which depends on $c_n$ only. I mean, informally you prove that that $A_nrightarrow e^{lim_n c_n}$.
      $endgroup$
      – Yanko
      Jan 7 at 22:08














      $begingroup$
      Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
      $endgroup$
      – amir na
      Jan 7 at 22:21




      $begingroup$
      Very good. It is a simple and clever proof for this problem which is also easy to remember when somebody asks for a proof. Thanks.
      $endgroup$
      – amir na
      Jan 7 at 22:21











      5












      $begingroup$

      If the limit does exist, we take logs of both sides, getting
      $$
      lim_{n to infty} (1+a_n)^{b_n}
      = lim_{n to infty} expleft(b_n ln (1+a_n)right)
      = exp left( lim_{n to infty} b_n ln (1+a_n)right)
      = exp left( lim_{n to infty} frac{ln (1+a_n)}{1/b_n}right)
      $$
      So if $a_n to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        If the limit does exist, we take logs of both sides, getting
        $$
        lim_{n to infty} (1+a_n)^{b_n}
        = lim_{n to infty} expleft(b_n ln (1+a_n)right)
        = exp left( lim_{n to infty} b_n ln (1+a_n)right)
        = exp left( lim_{n to infty} frac{ln (1+a_n)}{1/b_n}right)
        $$
        So if $a_n to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          If the limit does exist, we take logs of both sides, getting
          $$
          lim_{n to infty} (1+a_n)^{b_n}
          = lim_{n to infty} expleft(b_n ln (1+a_n)right)
          = exp left( lim_{n to infty} b_n ln (1+a_n)right)
          = exp left( lim_{n to infty} frac{ln (1+a_n)}{1/b_n}right)
          $$
          So if $a_n to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....






          share|cite|improve this answer









          $endgroup$



          If the limit does exist, we take logs of both sides, getting
          $$
          lim_{n to infty} (1+a_n)^{b_n}
          = lim_{n to infty} expleft(b_n ln (1+a_n)right)
          = exp left( lim_{n to infty} b_n ln (1+a_n)right)
          = exp left( lim_{n to infty} frac{ln (1+a_n)}{1/b_n}right)
          $$
          So if $a_n to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 21:49









          gt6989bgt6989b

          33.9k22455




          33.9k22455























              1












              $begingroup$

              Let $b_n,a_n$ such that $lim_n a_nb_n=c$. Moreover assume that $a_nrightarrow 0$.



              Then we have



              $$(1+a_n)^{b_n} = (1+a_n)^{frac{c}{a_n}} cdot (1+a_n)^{b_n-frac{c}{a_n}}$$



              We deal with each term alone:



              First, since $(1+frac{1}{n})^nrightarrow e$ we have that $(1+frac{1}{n})^{cn}rightarrow e^c$. As $a_nrightarrow 0$ we can substitute $a_n$ with $frac{1}{n}$.



              It is therefore enough to show that $(1+a_n)^{b_n-frac{c}{a_n}}rightarrow 0$.



              For every $varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<varepsilon$. This implies that $|b_n-frac{c}{a_n}|<frac{varepsilon}{a_n}$.



              Therefore $(1+a_n)^{b_n-frac{c}{a_n}} < (1+a_n)^{frac{varepsilon}{a_n}}$ and the latter convergence to $e^varepsilon$. As $varepsilon$ is arbitrary small, $e^varepsilon$ is arbitrary close to $1$ and this completes the proof.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $b_n,a_n$ such that $lim_n a_nb_n=c$. Moreover assume that $a_nrightarrow 0$.



                Then we have



                $$(1+a_n)^{b_n} = (1+a_n)^{frac{c}{a_n}} cdot (1+a_n)^{b_n-frac{c}{a_n}}$$



                We deal with each term alone:



                First, since $(1+frac{1}{n})^nrightarrow e$ we have that $(1+frac{1}{n})^{cn}rightarrow e^c$. As $a_nrightarrow 0$ we can substitute $a_n$ with $frac{1}{n}$.



                It is therefore enough to show that $(1+a_n)^{b_n-frac{c}{a_n}}rightarrow 0$.



                For every $varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<varepsilon$. This implies that $|b_n-frac{c}{a_n}|<frac{varepsilon}{a_n}$.



                Therefore $(1+a_n)^{b_n-frac{c}{a_n}} < (1+a_n)^{frac{varepsilon}{a_n}}$ and the latter convergence to $e^varepsilon$. As $varepsilon$ is arbitrary small, $e^varepsilon$ is arbitrary close to $1$ and this completes the proof.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $b_n,a_n$ such that $lim_n a_nb_n=c$. Moreover assume that $a_nrightarrow 0$.



                  Then we have



                  $$(1+a_n)^{b_n} = (1+a_n)^{frac{c}{a_n}} cdot (1+a_n)^{b_n-frac{c}{a_n}}$$



                  We deal with each term alone:



                  First, since $(1+frac{1}{n})^nrightarrow e$ we have that $(1+frac{1}{n})^{cn}rightarrow e^c$. As $a_nrightarrow 0$ we can substitute $a_n$ with $frac{1}{n}$.



                  It is therefore enough to show that $(1+a_n)^{b_n-frac{c}{a_n}}rightarrow 0$.



                  For every $varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<varepsilon$. This implies that $|b_n-frac{c}{a_n}|<frac{varepsilon}{a_n}$.



                  Therefore $(1+a_n)^{b_n-frac{c}{a_n}} < (1+a_n)^{frac{varepsilon}{a_n}}$ and the latter convergence to $e^varepsilon$. As $varepsilon$ is arbitrary small, $e^varepsilon$ is arbitrary close to $1$ and this completes the proof.






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                  $endgroup$



                  Let $b_n,a_n$ such that $lim_n a_nb_n=c$. Moreover assume that $a_nrightarrow 0$.



                  Then we have



                  $$(1+a_n)^{b_n} = (1+a_n)^{frac{c}{a_n}} cdot (1+a_n)^{b_n-frac{c}{a_n}}$$



                  We deal with each term alone:



                  First, since $(1+frac{1}{n})^nrightarrow e$ we have that $(1+frac{1}{n})^{cn}rightarrow e^c$. As $a_nrightarrow 0$ we can substitute $a_n$ with $frac{1}{n}$.



                  It is therefore enough to show that $(1+a_n)^{b_n-frac{c}{a_n}}rightarrow 0$.



                  For every $varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<varepsilon$. This implies that $|b_n-frac{c}{a_n}|<frac{varepsilon}{a_n}$.



                  Therefore $(1+a_n)^{b_n-frac{c}{a_n}} < (1+a_n)^{frac{varepsilon}{a_n}}$ and the latter convergence to $e^varepsilon$. As $varepsilon$ is arbitrary small, $e^varepsilon$ is arbitrary close to $1$ and this completes the proof.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 22:06









                  YankoYanko

                  6,4081528




                  6,4081528






























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