Prove the boundedness condition
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Suppose $a_ngeq 0.$ Then $sum_{n=1}^{infty}a_n$ converges if and only if the sequence of partial sums $(s_n)=(sum_{j=1}^na_j)$ is bounded. Use the axiom of completeness or one of its consequences to prove this result.
So the axiom of completeness says that there are no 'gaps' in the real number line. But I don't know where to USE the axiom of completeness
sequences-and-series
edited Nov 17 at 21:05
John B
12.2k51740
asked Nov 17 at 20:59
m.bazza
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Suppose $a_ngeq 0.$ Then $sum_{n=1}^{infty}a_n$ converges if and only if the sequence of partial sums $(s_n)=(sum_{j=1}^na_j)$ is bounded. Use the axiom of completeness or one of its consequences to prove this result.
So the axiom of completeness says that there are no 'gaps' in the real number line. But I don't know where to USE the axiom of completeness
sequences-and-series
edited Nov 17 at 21:05
John B
12.2k51740
asked Nov 17 at 20:59
m.bazza
827
$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
1
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $a_ngeq 0.$ Then $sum_{n=1}^{infty}a_n$ converges if and only if the sequence of partial sums $(s_n)=(sum_{j=1}^na_j)$ is bounded. Use the axiom of completeness or one of its consequences to prove this result.
So the axiom of completeness says that there are no 'gaps' in the real number line. But I don't know where to USE the axiom of completeness
sequences-and-series
edited Nov 17 at 21:05
John B
12.2k51740
asked Nov 17 at 20:59
m.bazza
827
Suppose $a_ngeq 0.$ Then $sum_{n=1}^{infty}a_n$ converges if and only if the sequence of partial sums $(s_n)=(sum_{j=1}^na_j)$ is bounded. Use the axiom of completeness or one of its consequences to prove this result.
So the axiom of completeness says that there are no 'gaps' in the real number line. But I don't know where to USE the axiom of completeness
sequences-and-series
sequences-and-series
edited Nov 17 at 21:05
John B
12.2k51740
asked Nov 17 at 20:59
m.bazza
827
edited Nov 17 at 21:05
John B
12.2k51740
asked Nov 17 at 20:59
m.bazza
827
edited Nov 17 at 21:05
John B
12.2k51740
edited Nov 17 at 21:05
John B
12.2k51740
edited Nov 17 at 21:05
John B
12.2k51740
12.2k51740
asked Nov 17 at 20:59
m.bazza
827
asked Nov 17 at 20:59
m.bazza
827
asked Nov 17 at 20:59
m.bazza
827
827
$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
1
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05
add a comment |
$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
1
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05
$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
1
1
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05
add a comment |
1 Answer
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The axiom of completeness implies that a nonempty set of real numbers which is bounded above has a least upper bound.
In your case the sequence of partial sum is nonempty and the supernum happens to be the limit.
You use definition of the least upper bound to prove your statement.
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The axiom of completeness implies that a nonempty set of real numbers which is bounded above has a least upper bound.
In your case the sequence of partial sum is nonempty and the supernum happens to be the limit.
You use definition of the least upper bound to prove your statement.
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
The axiom of completeness implies that a nonempty set of real numbers which is bounded above has a least upper bound.
In your case the sequence of partial sum is nonempty and the supernum happens to be the limit.
You use definition of the least upper bound to prove your statement.
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
The axiom of completeness implies that a nonempty set of real numbers which is bounded above has a least upper bound.
In your case the sequence of partial sum is nonempty and the supernum happens to be the limit.
You use definition of the least upper bound to prove your statement.
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
The axiom of completeness implies that a nonempty set of real numbers which is bounded above has a least upper bound.
In your case the sequence of partial sum is nonempty and the supernum happens to be the limit.
You use definition of the least upper bound to prove your statement.
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
answered Nov 17 at 21:07
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
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$a_ngeq 0$ as stated in the question <--- a reply to a deleted comment
– m.bazza
Nov 17 at 21:02
1
The first step towards using something is to state it. Not the thing you were told it kind-of means, not the thing you think it means: the actual statement.
– Saucy O'Path
Nov 17 at 21:03
The axiom is: Every non-empty subset of the reals that is bounded above has a least upper bound.
– m.bazza
Nov 17 at 21:05