Show that the sequence $a_n = (1 + frac{1}{n})^n$ is bounded. [duplicate]
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Proving $mathrm e <3$
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Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.
real-analysis sequences-and-series convergence upper-lower-bounds
edited Nov 25 at 10:21
Rebellos
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Nov 25 at 10:51
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This question already has an answer here:
Proving $mathrm e <3$
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Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.
real-analysis sequences-and-series convergence upper-lower-bounds
edited Nov 25 at 10:21
Rebellos
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Nov 25 at 10:51
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.
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Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30
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up vote
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This question already has an answer here:
Proving $mathrm e <3$
7 answers
Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.
real-analysis sequences-and-series convergence upper-lower-bounds
edited Nov 25 at 10:21
Rebellos
13.2k21142
asked Nov 25 at 10:15
user619263
193
This question already has an answer here:
Proving $mathrm e <3$
7 answers
Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.
This question already has an answer here:
Proving $mathrm e <3$
7 answers
real-analysis sequences-and-series convergence upper-lower-bounds
real-analysis sequences-and-series convergence upper-lower-bounds
edited Nov 25 at 10:21
Rebellos
13.2k21142
asked Nov 25 at 10:15
user619263
193
edited Nov 25 at 10:21
Rebellos
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asked Nov 25 at 10:15
user619263
193
edited Nov 25 at 10:21
Rebellos
13.2k21142
edited Nov 25 at 10:21
Rebellos
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edited Nov 25 at 10:21
Rebellos
13.2k21142
13.2k21142
asked Nov 25 at 10:15
user619263
193
asked Nov 25 at 10:15
user619263
193
asked Nov 25 at 10:15
user619263
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Nov 25 at 10:51
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.
3
Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30
add a comment |
3
Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30
3
3
Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30
add a comment |
2 Answers
2
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oldest
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up vote
0
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accepted
Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$
and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
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up vote
0
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Hints:
$$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$
$$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$
Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
answered Nov 25 at 10:51
KM101
3,494417
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$
and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
add a comment |
up vote
0
down vote
accepted
Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$
and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$
and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$
and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 10:47
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
add a comment |
add a comment |
up vote
0
down vote
Hints:
$$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$
$$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$
Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
answered Nov 25 at 10:51
KM101
3,494417
add a comment |
up vote
0
down vote
Hints:
$$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$
$$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$
Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
answered Nov 25 at 10:51
KM101
3,494417
add a comment |
up vote
0
down vote
up vote
0
down vote
Hints:
$$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$
$$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$
Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
answered Nov 25 at 10:51
KM101
3,494417
Hints:
$$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$
$$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$
Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)
answered Nov 25 at 10:51
KM101
3,494417
answered Nov 25 at 10:51
KM101
3,494417
answered Nov 25 at 10:51
KM101
3,494417
answered Nov 25 at 10:51
KM101
3,494417
3,494417
add a comment |
add a comment |
3
Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 at 10:22
Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 at 10:30