Prove that the sequence (1+1/n)^n is convergent [duplicate] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?Is this proof that $lim_n to infty (1+1/n)^n$ exists (1) new, (2) interesting?Proving convergence of a sequence given by $a_n+1=A+Ba_n^3$Show that the sequence $x_n=left[1+frac(-1)^nnright]$ is convergent using monotone convergence theorem.Show that the sequence $x_n+1=x_n(2-x_n)$ is convergentIf $sum a_n$ is convergent and $a_n>0$, then $sum (-1)^n a_n$ is convergent.The Limit of The Sequence Given By a Recursive RelationLet $x_1 = 2$, $x_n+1=2+frac1x_n$, $n in mathbbN$. Does the sequence converge or diverge?Prove Bernoulli's inequality for irrationalsMonotone of a sequence without Bernouli's InequalityConvergent Sequence Summation ProblemAbout the proof that “ a bounded and monotone sequence is convergent”
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Prove that the sequence (1+1/n)^n is convergent [duplicate]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?Is this proof that $lim_n to infty (1+1/n)^n$ exists (1) new, (2) interesting?Proving convergence of a sequence given by $a_n+1=A+Ba_n^3$Show that the sequence $x_n=left[1+frac(-1)^nnright]$ is convergent using monotone convergence theorem.Show that the sequence $x_n+1=x_n(2-x_n)$ is convergentIf $sum a_n$ is convergent and $a_n>0$, then $sum (-1)^n a_n$ is convergent.The Limit of The Sequence Given By a Recursive RelationLet $x_1 = 2$, $x_n+1=2+frac1x_n$, $n in mathbbN$. Does the sequence converge or diverge?Prove Bernoulli's inequality for irrationalsMonotone of a sequence without Bernouli's InequalityConvergent Sequence Summation ProblemAbout the proof that “ a bounded and monotone sequence is convergent”
$begingroup$
This question already has an answer here:
What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?
3 answers
I know the proof using binomial expansion and then by monotone convergence theorem.
But i want to collect some other proofs without using the binomial expansion.
*if you could provide the answer without the concept of calculus it will be appreciated *
But any way do provide whatever method you can.
sequences-and-series
asked Mar 27 at 17:05
risav jainrisav jain
94
$endgroup$
marked as duplicate by Somos, Sil, YiFan, Shailesh, Eevee Trainer Mar 28 at 0:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?
3 answers
I know the proof using binomial expansion and then by monotone convergence theorem.
But i want to collect some other proofs without using the binomial expansion.
*if you could provide the answer without the concept of calculus it will be appreciated *
But any way do provide whatever method you can.
sequences-and-series
asked Mar 27 at 17:05
risav jainrisav jain
94
$endgroup$
marked as duplicate by Somos, Sil, YiFan, Shailesh, Eevee Trainer Mar 28 at 0:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
$endgroup$
– Misha Lavrov
Mar 27 at 17:24
$begingroup$
math.stackexchange.com/questions/389793/…
$endgroup$
– marty cohen
Mar 27 at 17:56
add a comment |
$begingroup$
This question already has an answer here:
What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?
3 answers
I know the proof using binomial expansion and then by monotone convergence theorem.
But i want to collect some other proofs without using the binomial expansion.
*if you could provide the answer without the concept of calculus it will be appreciated *
But any way do provide whatever method you can.
sequences-and-series
asked Mar 27 at 17:05
risav jainrisav jain
94
$endgroup$
This question already has an answer here:
What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?
3 answers
I know the proof using binomial expansion and then by monotone convergence theorem.
But i want to collect some other proofs without using the binomial expansion.
*if you could provide the answer without the concept of calculus it will be appreciated *
But any way do provide whatever method you can.
This question already has an answer here:
What is the most elementary proof that $lim_n to infty (1+1/n)^n$ exists?
3 answers
sequences-and-series
sequences-and-series
asked Mar 27 at 17:05
risav jainrisav jain
94
asked Mar 27 at 17:05
risav jainrisav jain
94
edited Mar 27 at 17:19
risav jain
asked Mar 27 at 17:05
risav jainrisav jain
94
asked Mar 27 at 17:05
risav jainrisav jain
94
asked Mar 27 at 17:05
risav jainrisav jain
94
94
marked as duplicate by Somos, Sil, YiFan, Shailesh, Eevee Trainer Mar 28 at 0:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Somos, Sil, YiFan, Shailesh, Eevee Trainer Mar 28 at 0:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
$endgroup$
– Misha Lavrov
Mar 27 at 17:24
$begingroup$
math.stackexchange.com/questions/389793/…
$endgroup$
– marty cohen
Mar 27 at 17:56
add a comment |
1
$begingroup$
It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
$endgroup$
– Misha Lavrov
Mar 27 at 17:24
$begingroup$
math.stackexchange.com/questions/389793/…
$endgroup$
– marty cohen
Mar 27 at 17:56
1
1
$begingroup$
It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
$endgroup$
– Misha Lavrov
Mar 27 at 17:24
$begingroup$
It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
$endgroup$
– Misha Lavrov
Mar 27 at 17:24
$begingroup$
math.stackexchange.com/questions/389793/…
$endgroup$
– marty cohen
Mar 27 at 17:56
$begingroup$
math.stackexchange.com/questions/389793/…
$endgroup$
– marty cohen
Mar 27 at 17:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If $a_n = left(1 + frac1nright)^n$, then $ln(a_n) = n cdot lnleft(1+frac1nright)$. This is indeterminate, but you can apply l'Hopital's rule to find $ln(a_n)to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_nto e$.
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
add a comment |
$begingroup$
Let $a_n$ the given sequence. We have
$$w_n:=ln(a_n+1)-ln a_n=lnfraca_n+1a_n=(n+1)ln(1+1/(n+1))-nln(1+1/n)=O(frac1n^2)$$
so the series $sum w_n$ is convergent and so by telescoping the sequence $(ln(a_n))$ is convergent. Conclude using the exponential function.
answered Mar 27 at 17:23
user296113user296113
7,015928
$endgroup$
add a comment |
$begingroup$
You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+frac1n)^n$.
- Show that $a_nleqsum_k=0^nfrac1k!$ for all $ninmathbb N$.
- Show that $a_n$ is monotonically increasing.
To show (2), you have to prove that $fraca_n+1a_ngeq1$; some calculations and Bernoulli's inequality are involved here:
$$fraca_n+1a_n=fracleft(1+frac11+nright)^n+1left(1+frac1nright)^n=left(1+frac1nright)left(frac1+frac11+n1+frac1nright)^n+1=left(1+frac1nright)left(fracn^2+2nn^2+2n+1right)^n+1=left(1+frac1nright)left(1-frac1n^2+2n+1right)^n+1geqleft(1+frac1nright)left(1-fracn+1n^2+2n+1right)=fracn+1nfracnn+1=1$$
Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.
answered Mar 27 at 17:27
st.mathst.math
1,268115
$endgroup$
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
$endgroup$
– marty cohen
Mar 27 at 21:13
$begingroup$
You are right, thanks. :) Corrected it.
$endgroup$
– st.math
Mar 27 at 22:02
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a_n = left(1 + frac1nright)^n$, then $ln(a_n) = n cdot lnleft(1+frac1nright)$. This is indeterminate, but you can apply l'Hopital's rule to find $ln(a_n)to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_nto e$.
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
add a comment |
$begingroup$
If $a_n = left(1 + frac1nright)^n$, then $ln(a_n) = n cdot lnleft(1+frac1nright)$. This is indeterminate, but you can apply l'Hopital's rule to find $ln(a_n)to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_nto e$.
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
add a comment |
$begingroup$
If $a_n = left(1 + frac1nright)^n$, then $ln(a_n) = n cdot lnleft(1+frac1nright)$. This is indeterminate, but you can apply l'Hopital's rule to find $ln(a_n)to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_nto e$.
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
$endgroup$
If $a_n = left(1 + frac1nright)^n$, then $ln(a_n) = n cdot lnleft(1+frac1nright)$. This is indeterminate, but you can apply l'Hopital's rule to find $ln(a_n)to 1$ with $n$, so that, using the fact that continuous functions preserve limits, $a_nto e$.
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
answered Mar 27 at 17:14
MatthewPeterMatthewPeter
191
191
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
add a comment |
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
$begingroup$
Is their a way without using the concept of calculus?
$endgroup$
– risav jain
Mar 27 at 17:16
add a comment |
$begingroup$
Let $a_n$ the given sequence. We have
$$w_n:=ln(a_n+1)-ln a_n=lnfraca_n+1a_n=(n+1)ln(1+1/(n+1))-nln(1+1/n)=O(frac1n^2)$$
so the series $sum w_n$ is convergent and so by telescoping the sequence $(ln(a_n))$ is convergent. Conclude using the exponential function.
answered Mar 27 at 17:23
user296113user296113
7,015928
$endgroup$
add a comment |
$begingroup$
Let $a_n$ the given sequence. We have
$$w_n:=ln(a_n+1)-ln a_n=lnfraca_n+1a_n=(n+1)ln(1+1/(n+1))-nln(1+1/n)=O(frac1n^2)$$
so the series $sum w_n$ is convergent and so by telescoping the sequence $(ln(a_n))$ is convergent. Conclude using the exponential function.
answered Mar 27 at 17:23
user296113user296113
7,015928
$endgroup$
add a comment |
$begingroup$
Let $a_n$ the given sequence. We have
$$w_n:=ln(a_n+1)-ln a_n=lnfraca_n+1a_n=(n+1)ln(1+1/(n+1))-nln(1+1/n)=O(frac1n^2)$$
so the series $sum w_n$ is convergent and so by telescoping the sequence $(ln(a_n))$ is convergent. Conclude using the exponential function.
answered Mar 27 at 17:23
user296113user296113
7,015928
$endgroup$
Let $a_n$ the given sequence. We have
$$w_n:=ln(a_n+1)-ln a_n=lnfraca_n+1a_n=(n+1)ln(1+1/(n+1))-nln(1+1/n)=O(frac1n^2)$$
so the series $sum w_n$ is convergent and so by telescoping the sequence $(ln(a_n))$ is convergent. Conclude using the exponential function.
answered Mar 27 at 17:23
user296113user296113
7,015928
answered Mar 27 at 17:23
user296113user296113
7,015928
answered Mar 27 at 17:23
user296113user296113
7,015928
answered Mar 27 at 17:23
user296113user296113
7,015928
7,015928
add a comment |
add a comment |
$begingroup$
You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+frac1n)^n$.
- Show that $a_nleqsum_k=0^nfrac1k!$ for all $ninmathbb N$.
- Show that $a_n$ is monotonically increasing.
To show (2), you have to prove that $fraca_n+1a_ngeq1$; some calculations and Bernoulli's inequality are involved here:
$$fraca_n+1a_n=fracleft(1+frac11+nright)^n+1left(1+frac1nright)^n=left(1+frac1nright)left(frac1+frac11+n1+frac1nright)^n+1=left(1+frac1nright)left(fracn^2+2nn^2+2n+1right)^n+1=left(1+frac1nright)left(1-frac1n^2+2n+1right)^n+1geqleft(1+frac1nright)left(1-fracn+1n^2+2n+1right)=fracn+1nfracnn+1=1$$
Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.
answered Mar 27 at 17:27
st.mathst.math
1,268115
$endgroup$
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
$endgroup$
– marty cohen
Mar 27 at 21:13
$begingroup$
You are right, thanks. :) Corrected it.
$endgroup$
– st.math
Mar 27 at 22:02
add a comment |
$begingroup$
You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+frac1n)^n$.
- Show that $a_nleqsum_k=0^nfrac1k!$ for all $ninmathbb N$.
- Show that $a_n$ is monotonically increasing.
To show (2), you have to prove that $fraca_n+1a_ngeq1$; some calculations and Bernoulli's inequality are involved here:
$$fraca_n+1a_n=fracleft(1+frac11+nright)^n+1left(1+frac1nright)^n=left(1+frac1nright)left(frac1+frac11+n1+frac1nright)^n+1=left(1+frac1nright)left(fracn^2+2nn^2+2n+1right)^n+1=left(1+frac1nright)left(1-frac1n^2+2n+1right)^n+1geqleft(1+frac1nright)left(1-fracn+1n^2+2n+1right)=fracn+1nfracnn+1=1$$
Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.
answered Mar 27 at 17:27
st.mathst.math
1,268115
$endgroup$
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
$endgroup$
– marty cohen
Mar 27 at 21:13
$begingroup$
You are right, thanks. :) Corrected it.
$endgroup$
– st.math
Mar 27 at 22:02
add a comment |
$begingroup$
You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+frac1n)^n$.
- Show that $a_nleqsum_k=0^nfrac1k!$ for all $ninmathbb N$.
- Show that $a_n$ is monotonically increasing.
To show (2), you have to prove that $fraca_n+1a_ngeq1$; some calculations and Bernoulli's inequality are involved here:
$$fraca_n+1a_n=fracleft(1+frac11+nright)^n+1left(1+frac1nright)^n=left(1+frac1nright)left(frac1+frac11+n1+frac1nright)^n+1=left(1+frac1nright)left(fracn^2+2nn^2+2n+1right)^n+1=left(1+frac1nright)left(1-frac1n^2+2n+1right)^n+1geqleft(1+frac1nright)left(1-fracn+1n^2+2n+1right)=fracn+1nfracnn+1=1$$
Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.
answered Mar 27 at 17:27
st.mathst.math
1,268115
$endgroup$
You can do this in two steps; here, no calculus is needed:
Let $a_n=(1+frac1n)^n$.
- Show that $a_nleqsum_k=0^nfrac1k!$ for all $ninmathbb N$.
- Show that $a_n$ is monotonically increasing.
To show (2), you have to prove that $fraca_n+1a_ngeq1$; some calculations and Bernoulli's inequality are involved here:
$$fraca_n+1a_n=fracleft(1+frac11+nright)^n+1left(1+frac1nright)^n=left(1+frac1nright)left(frac1+frac11+n1+frac1nright)^n+1=left(1+frac1nright)left(fracn^2+2nn^2+2n+1right)^n+1=left(1+frac1nright)left(1-frac1n^2+2n+1right)^n+1geqleft(1+frac1nright)left(1-fracn+1n^2+2n+1right)=fracn+1nfracnn+1=1$$
Because the series on the right hand side in (1) is convergent, $(a_n)$ is bounded. Together with (2), this implies that the sequence converges.
answered Mar 27 at 17:27
st.mathst.math
1,268115
edited Mar 27 at 21:55
answered Mar 27 at 17:27
st.mathst.math
1,268115
answered Mar 27 at 17:27
st.mathst.math
1,268115
answered Mar 27 at 17:27
st.mathst.math
1,268115
1,268115
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
$endgroup$
– marty cohen
Mar 27 at 21:13
$begingroup$
You are right, thanks. :) Corrected it.
$endgroup$
– st.math
Mar 27 at 22:02
add a comment |
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
$endgroup$
– marty cohen
Mar 27 at 21:13
$begingroup$
You are right, thanks. :) Corrected it.
$endgroup$
– st.math
Mar 27 at 22:02
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
1 is easy, 2 is not.
$endgroup$
– marty cohen
Mar 27 at 17:55
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Edited my answer to include the calculations for (2).
$endgroup$
– st.math
Mar 27 at 18:08
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
$endgroup$
– marty cohen
Mar 27 at 21:09
$begingroup$
Third line, second expression should be 1-, not 1+, so it's $1-1/(n+1) = n/(n+1)$ so you get 1.
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– marty cohen
Mar 27 at 21:09
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See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
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– marty cohen
Mar 27 at 21:13
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See this question of mine which does essentially the same thing in more detail: math.stackexchange.com/questions/1387272/…
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– marty cohen
Mar 27 at 21:13
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You are right, thanks. :) Corrected it.
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– st.math
Mar 27 at 22:02
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You are right, thanks. :) Corrected it.
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– st.math
Mar 27 at 22:02
add a comment |
aF1W,l3,00nlU8x EC,Yzco84lZxPVh85a,hyHBncM
1
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It is hard to avoid "the concept of calculus" since limits and convergent sequences are a part of that concept. On the other hand, it would help to specify what tools you're happy with using, since this result is used in developing some of them. (For example, if you define $e^x = lim_n to infty (1 + x/n)^n$, then clearly we should not be using $e^x$ in the process of showing that this limit exists.)
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– Misha Lavrov
Mar 27 at 17:24
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math.stackexchange.com/questions/389793/…
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– marty cohen
Mar 27 at 17:56