a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$
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I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:
$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$
but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!
I am really confused!
real-analysis sequences-and-series
add a comment |
up vote
-1
down vote
favorite
I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:
$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$
but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!
I am really confused!
real-analysis sequences-and-series
the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
1
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:
$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$
but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!
I am really confused!
real-analysis sequences-and-series
I have got a sequence $frac19,frac{7}{17},frac{17}{25}, cdots$ and got two different version for general term:
$$frac{2n^2-1}{8n+1} & frac{8(n-1)+(-1)^{n+1}}{8n+1}$$
but uisng the 1st one we get the sequence is divergent and the second one gives us convergence!!
I am really confused!
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Nov 17 at 3:01
user8795
5,57961846
5,57961846
the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
1
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12
add a comment |
the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
1
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12
the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
1
1
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
add a comment |
up vote
0
down vote
This is possible, finitely many terms do not tell us about the convergence of the whole sequence.
Also, in general, we can't tell the whole sequence from finitely many terms.
add a comment |
up vote
0
down vote
You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.
You only have three terms of your sequence, depending on what happens with other terms everything is possible.
For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
add a comment |
up vote
2
down vote
These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
add a comment |
up vote
2
down vote
up vote
2
down vote
These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.
These two expressions only agree until $n=3$. At $n=4$, the first expression gives $frac{31}{33}$, whereas the second gives $frac{23}{33}$. As $n to infty$, these expressions are giving you two different sequences. So one will converge and the other won't but their behaviors don't have to be dependent on each other just because the first few terms are a match.
answered Nov 17 at 3:06
Anurag A
24.9k12249
24.9k12249
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
add a comment |
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
1
1
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
When you are given a few terms of a sequence there is no mathematical reason to assume the pattern continues. The sequence could do anything at all. We generally expect that the sequence will continue in the "obviously simplest" way and I believe the problem setter has the responsibility to make sure there is one "obviously simplest" way. OP has demonstrated to my satisfaction that the problem setter has failed. I think this answer is more responsive to the question than ones that just say the sequence can go anywhere (though they are correct)
– Ross Millikan
Nov 17 at 3:17
add a comment |
up vote
0
down vote
This is possible, finitely many terms do not tell us about the convergence of the whole sequence.
Also, in general, we can't tell the whole sequence from finitely many terms.
add a comment |
up vote
0
down vote
This is possible, finitely many terms do not tell us about the convergence of the whole sequence.
Also, in general, we can't tell the whole sequence from finitely many terms.
add a comment |
up vote
0
down vote
up vote
0
down vote
This is possible, finitely many terms do not tell us about the convergence of the whole sequence.
Also, in general, we can't tell the whole sequence from finitely many terms.
This is possible, finitely many terms do not tell us about the convergence of the whole sequence.
Also, in general, we can't tell the whole sequence from finitely many terms.
answered Nov 17 at 3:03
Siong Thye Goh
94.6k1462114
94.6k1462114
add a comment |
add a comment |
up vote
0
down vote
You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.
You only have three terms of your sequence, depending on what happens with other terms everything is possible.
For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.
add a comment |
up vote
0
down vote
You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.
You only have three terms of your sequence, depending on what happens with other terms everything is possible.
For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.
add a comment |
up vote
0
down vote
up vote
0
down vote
You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.
You only have three terms of your sequence, depending on what happens with other terms everything is possible.
For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.
You have made up your formulas based on just three terms. Others may come up with other formulas or just make up other terms for the rest of the sequence.
You only have three terms of your sequence, depending on what happens with other terms everything is possible.
For example you can define the rest of your sequence as $$4,5,6,...n,...$$ to make it divergent or $$ 0,0,0,...$$ or anything else.
answered Nov 17 at 3:13
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
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the question appeared in one of my exams, so I would write that we can't conclude from finitely many terms?? @AnuragA
– user8795
Nov 17 at 3:09
1
Yes. You can show that fourth terms do not match so these are possibly two different sequences.
– Anurag A
Nov 17 at 3:12