Prove that the sequence is bounded : [closed]
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How to mathematically prove that the sequence $ {a_n=dfrac{3}{3^n} }$ is bounded ?
real-analysis upper-lower-bounds
edited Nov 25 '18 at 16:08
Bernard
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
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closed as off-topic by Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen Nov 25 '18 at 13:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to mathematically prove that the sequence $ {a_n=dfrac{3}{3^n} }$ is bounded ?
real-analysis upper-lower-bounds
edited Nov 25 '18 at 16:08
Bernard
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
$endgroup$
closed as off-topic by Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen Nov 25 '18 at 13:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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For $n>1$ the sequence is bounded by $1$.
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– Yadati Kiran
Nov 25 '18 at 12:25
3
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Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
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– Martin R
Nov 25 '18 at 12:27
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Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37
add a comment |
$begingroup$
How to mathematically prove that the sequence $ {a_n=dfrac{3}{3^n} }$ is bounded ?
real-analysis upper-lower-bounds
edited Nov 25 '18 at 16:08
Bernard
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
$endgroup$
How to mathematically prove that the sequence $ {a_n=dfrac{3}{3^n} }$ is bounded ?
real-analysis upper-lower-bounds
real-analysis upper-lower-bounds
edited Nov 25 '18 at 16:08
Bernard
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
edited Nov 25 '18 at 16:08
Bernard
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
edited Nov 25 '18 at 16:08
Bernard
119k639113
edited Nov 25 '18 at 16:08
Bernard
119k639113
edited Nov 25 '18 at 16:08
Bernard
119k639113
119k639113
asked Nov 25 '18 at 12:24
user619263user619263
143
asked Nov 25 '18 at 12:24
user619263user619263
143
asked Nov 25 '18 at 12:24
user619263user619263
143
143
closed as off-topic by Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen Nov 25 '18 at 13:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen Nov 25 '18 at 13:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Saad, José Carlos Santos, amWhy, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
For $n>1$ the sequence is bounded by $1$.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 12:25
3
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
$endgroup$
– Martin R
Nov 25 '18 at 12:27
$begingroup$
Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37
add a comment |
1
$begingroup$
For $n>1$ the sequence is bounded by $1$.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 12:25
3
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
$endgroup$
– Martin R
Nov 25 '18 at 12:27
$begingroup$
Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37
1
1
$begingroup$
For $n>1$ the sequence is bounded by $1$.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 12:25
$begingroup$
For $n>1$ the sequence is bounded by $1$.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 12:25
3
3
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
$endgroup$
– Martin R
Nov 25 '18 at 12:27
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
$endgroup$
– Martin R
Nov 25 '18 at 12:27
$begingroup$
Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37
$begingroup$
Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37
add a comment |
1 Answer
1
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$begingroup$
Given $n^{th}$ term is $ a_n=frac{3}{3^{n}}=frac{1}{3^{n-1}}$.
Now check that,
$ a_1=1> a_2=frac{1}{3}> a_3=frac{1}{3^2}> cdots >a_n=frac{1}{3^{n-1}}> cdots$
Thus clearly $ a_1=1$ is the upper bound of the sequence $ {frac{3}{3^n} }$.
Since the sequence ${frac{3}{3^n} }$ is decreasing, the lower bound of this sequence is the limit as follows:
$ lim_{n to infty} frac{3}{3^n}=lim_{n to infty} frac{1}{3^{n-1}} =0$.
Thus,
$ 0 < a_n leq 1 $.
Hence the given sequence is bounded.
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given $n^{th}$ term is $ a_n=frac{3}{3^{n}}=frac{1}{3^{n-1}}$.
Now check that,
$ a_1=1> a_2=frac{1}{3}> a_3=frac{1}{3^2}> cdots >a_n=frac{1}{3^{n-1}}> cdots$
Thus clearly $ a_1=1$ is the upper bound of the sequence $ {frac{3}{3^n} }$.
Since the sequence ${frac{3}{3^n} }$ is decreasing, the lower bound of this sequence is the limit as follows:
$ lim_{n to infty} frac{3}{3^n}=lim_{n to infty} frac{1}{3^{n-1}} =0$.
Thus,
$ 0 < a_n leq 1 $.
Hence the given sequence is bounded.
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
$endgroup$
add a comment |
$begingroup$
Given $n^{th}$ term is $ a_n=frac{3}{3^{n}}=frac{1}{3^{n-1}}$.
Now check that,
$ a_1=1> a_2=frac{1}{3}> a_3=frac{1}{3^2}> cdots >a_n=frac{1}{3^{n-1}}> cdots$
Thus clearly $ a_1=1$ is the upper bound of the sequence $ {frac{3}{3^n} }$.
Since the sequence ${frac{3}{3^n} }$ is decreasing, the lower bound of this sequence is the limit as follows:
$ lim_{n to infty} frac{3}{3^n}=lim_{n to infty} frac{1}{3^{n-1}} =0$.
Thus,
$ 0 < a_n leq 1 $.
Hence the given sequence is bounded.
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
$endgroup$
add a comment |
$begingroup$
Given $n^{th}$ term is $ a_n=frac{3}{3^{n}}=frac{1}{3^{n-1}}$.
Now check that,
$ a_1=1> a_2=frac{1}{3}> a_3=frac{1}{3^2}> cdots >a_n=frac{1}{3^{n-1}}> cdots$
Thus clearly $ a_1=1$ is the upper bound of the sequence $ {frac{3}{3^n} }$.
Since the sequence ${frac{3}{3^n} }$ is decreasing, the lower bound of this sequence is the limit as follows:
$ lim_{n to infty} frac{3}{3^n}=lim_{n to infty} frac{1}{3^{n-1}} =0$.
Thus,
$ 0 < a_n leq 1 $.
Hence the given sequence is bounded.
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
$endgroup$
Given $n^{th}$ term is $ a_n=frac{3}{3^{n}}=frac{1}{3^{n-1}}$.
Now check that,
$ a_1=1> a_2=frac{1}{3}> a_3=frac{1}{3^2}> cdots >a_n=frac{1}{3^{n-1}}> cdots$
Thus clearly $ a_1=1$ is the upper bound of the sequence $ {frac{3}{3^n} }$.
Since the sequence ${frac{3}{3^n} }$ is decreasing, the lower bound of this sequence is the limit as follows:
$ lim_{n to infty} frac{3}{3^n}=lim_{n to infty} frac{1}{3^{n-1}} =0$.
Thus,
$ 0 < a_n leq 1 $.
Hence the given sequence is bounded.
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
answered Nov 25 '18 at 12:38
M. A. SARKARM. A. SARKAR
2,1641619
2,1641619
add a comment |
add a comment |
1
$begingroup$
For $n>1$ the sequence is bounded by $1$.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 12:25
3
$begingroup$
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers
$endgroup$
– Martin R
Nov 25 '18 at 12:27
$begingroup$
Simply note that $0< a_{n+1}<a_nleq 3/a_{n_0}$, so this is a decreasing sequence bounded below and above.
$endgroup$
– MPW
Nov 25 '18 at 12:37