Mixing
General formula of $lim_{n to infty} (1+a_n)^{b_n}$ where $b_n to infty$
$begingroup$
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
edited Jan 7 at 21:44
gt6989b
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
$endgroup$
add a comment |
$begingroup$
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
edited Jan 7 at 21:44
gt6989b
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 7 at 21:44
gt6989b
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
$endgroup$
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
edited Jan 7 at 21:44
gt6989b
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
edited Jan 7 at 21:44
gt6989b
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
edited Jan 7 at 21:44
gt6989b
33.9k22455
edited Jan 7 at 21:44
gt6989b
33.9k22455
edited Jan 7 at 21:44
gt6989b
33.9k22455
33.9k22455
asked Jan 7 at 21:42
amir naamir na
625
asked Jan 7 at 21:42
amir naamir na
625
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $A_n$.
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
$endgroup$
$begingroup$
(+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
add a comment |
$begingroup$
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 $n$....
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 7 at 22:06
YankoYanko
6,5571529
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $A_n$.
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
$endgroup$
$begingroup$
(+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
add a comment |
$begingroup$
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 $A_n$.
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
$endgroup$
$begingroup$
(+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
add a comment |
$begingroup$
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 $A_n$.
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
$endgroup$
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 $A_n$.
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
edited Jan 7 at 22:10
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
answered Jan 7 at 22:06
SomeStrangeUserSomeStrangeUser
1,6501025
1,6501025
$begingroup$
(+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
add a comment |
$begingroup$
(+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
add a comment |
$begingroup$
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 $n$....
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
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 $n$....
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
$endgroup$
add a comment |
$begingroup$
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 $n$....
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
$endgroup$
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 $n$....
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
answered Jan 7 at 21:49
gt6989bgt6989b
33.9k22455
33.9k22455
add a comment |
add a comment |
$begingroup$
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.
answered Jan 7 at 22:06
YankoYanko
6,5571529
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 7 at 22:06
YankoYanko
6,5571529
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 7 at 22:06
YankoYanko
6,5571529
$endgroup$
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.
answered Jan 7 at 22:06
YankoYanko
6,5571529
answered Jan 7 at 22:06
YankoYanko
6,5571529
answered Jan 7 at 22:06
YankoYanko
6,5571529
answered Jan 7 at 22:06
YankoYanko
6,5571529
6,5571529
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$begingroup$
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