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What's wrong with the following proof that rearrangement of any convergent series (even a conditionally convergent one) converges
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$sum_n=1^infty a_n$ converges to $l$. Let $b_n$ be any rearrangement of $a_n$'s. We show that $sum_n=1^infty b_n$ also converges.
Let $s_k = sum_n=1^k a_n$, and $t_k = sum_n=1^k b_n$ be partial sums. Since for any $epsilon > 0$, there exists a number $N$ such that $forall n geq N$, $|s_n - l| < epsilon$. Or, equivalently $|s_m - s_N| < epsilon$ for any $m>N$. (also called Cauchy criterion)
Now, in the rearrangement series, choose an $M$ such that $a_1, a_2, dots, a_N$ are covered in $b_1, b_2, dots, b_M$. Therefore $|t_M - s_N| = |a_i_1 + a_i_2 + dots + a_i_M-N| leq |s_M' - s_N|$ for some $M' > N$. But since $|s_m - s_N| < epsilon$ for any $m>N$, we have $|t_M - s_N| < epsilon$. Consequently, $|t_M - l| < 2epsilon$. And the convergence of rearrangement follows.
sequences-and-series
asked Jul 22 at 16:58
student
62
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$sum_n=1^infty a_n$ converges to $l$. Let $b_n$ be any rearrangement of $a_n$'s. We show that $sum_n=1^infty b_n$ also converges.
Let $s_k = sum_n=1^k a_n$, and $t_k = sum_n=1^k b_n$ be partial sums. Since for any $epsilon > 0$, there exists a number $N$ such that $forall n geq N$, $|s_n - l| < epsilon$. Or, equivalently $|s_m - s_N| < epsilon$ for any $m>N$. (also called Cauchy criterion)
Now, in the rearrangement series, choose an $M$ such that $a_1, a_2, dots, a_N$ are covered in $b_1, b_2, dots, b_M$. Therefore $|t_M - s_N| = |a_i_1 + a_i_2 + dots + a_i_M-N| leq |s_M' - s_N|$ for some $M' > N$. But since $|s_m - s_N| < epsilon$ for any $m>N$, we have $|t_M - s_N| < epsilon$. Consequently, $|t_M - l| < 2epsilon$. And the convergence of rearrangement follows.
sequences-and-series
asked Jul 22 at 16:58
student
62
3
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
Gotcha! Thanks.
– student
Jul 22 at 18:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$sum_n=1^infty a_n$ converges to $l$. Let $b_n$ be any rearrangement of $a_n$'s. We show that $sum_n=1^infty b_n$ also converges.
Let $s_k = sum_n=1^k a_n$, and $t_k = sum_n=1^k b_n$ be partial sums. Since for any $epsilon > 0$, there exists a number $N$ such that $forall n geq N$, $|s_n - l| < epsilon$. Or, equivalently $|s_m - s_N| < epsilon$ for any $m>N$. (also called Cauchy criterion)
Now, in the rearrangement series, choose an $M$ such that $a_1, a_2, dots, a_N$ are covered in $b_1, b_2, dots, b_M$. Therefore $|t_M - s_N| = |a_i_1 + a_i_2 + dots + a_i_M-N| leq |s_M' - s_N|$ for some $M' > N$. But since $|s_m - s_N| < epsilon$ for any $m>N$, we have $|t_M - s_N| < epsilon$. Consequently, $|t_M - l| < 2epsilon$. And the convergence of rearrangement follows.
sequences-and-series
asked Jul 22 at 16:58
student
62
$sum_n=1^infty a_n$ converges to $l$. Let $b_n$ be any rearrangement of $a_n$'s. We show that $sum_n=1^infty b_n$ also converges.
Let $s_k = sum_n=1^k a_n$, and $t_k = sum_n=1^k b_n$ be partial sums. Since for any $epsilon > 0$, there exists a number $N$ such that $forall n geq N$, $|s_n - l| < epsilon$. Or, equivalently $|s_m - s_N| < epsilon$ for any $m>N$. (also called Cauchy criterion)
Now, in the rearrangement series, choose an $M$ such that $a_1, a_2, dots, a_N$ are covered in $b_1, b_2, dots, b_M$. Therefore $|t_M - s_N| = |a_i_1 + a_i_2 + dots + a_i_M-N| leq |s_M' - s_N|$ for some $M' > N$. But since $|s_m - s_N| < epsilon$ for any $m>N$, we have $|t_M - s_N| < epsilon$. Consequently, $|t_M - l| < 2epsilon$. And the convergence of rearrangement follows.
sequences-and-series
asked Jul 22 at 16:58
student
62
asked Jul 22 at 16:58
student
62
asked Jul 22 at 16:58
student
62
asked Jul 22 at 16:58
student
62
62
3
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
Gotcha! Thanks.
– student
Jul 22 at 18:51
add a comment |Â
3
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
Gotcha! Thanks.
– student
Jul 22 at 18:51
3
3
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
Gotcha! Thanks.
– student
Jul 22 at 18:51
Gotcha! Thanks.
– student
Jul 22 at 18:51
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:
$$
|a_i_1 + a_i_2 + ... + a_i_M-N| leq |s_M' - s_N| =
|a_N+1 + a_N+2 + ... + a_M'|
$$
The indices $i_1, ..., i_M-N$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:
$$
|1+1| nleq |1+(-1)+1|
$$
answered Jul 22 at 17:10
Sambo
1,2561427
Thanks @Sambo .
– student
Jul 22 at 18:51
add a comment |Â
up vote
2
down vote
Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_i_1+dots|$ should be $|b_i_1+dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:
$$
|a_i_1 + a_i_2 + ... + a_i_M-N| leq |s_M' - s_N| =
|a_N+1 + a_N+2 + ... + a_M'|
$$
The indices $i_1, ..., i_M-N$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:
$$
|1+1| nleq |1+(-1)+1|
$$
answered Jul 22 at 17:10
Sambo
1,2561427
Thanks @Sambo .
– student
Jul 22 at 18:51
add a comment |Â
up vote
2
down vote
As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:
$$
|a_i_1 + a_i_2 + ... + a_i_M-N| leq |s_M' - s_N| =
|a_N+1 + a_N+2 + ... + a_M'|
$$
The indices $i_1, ..., i_M-N$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:
$$
|1+1| nleq |1+(-1)+1|
$$
answered Jul 22 at 17:10
Sambo
1,2561427
Thanks @Sambo .
– student
Jul 22 at 18:51
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:
$$
|a_i_1 + a_i_2 + ... + a_i_M-N| leq |s_M' - s_N| =
|a_N+1 + a_N+2 + ... + a_M'|
$$
The indices $i_1, ..., i_M-N$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:
$$
|1+1| nleq |1+(-1)+1|
$$
answered Jul 22 at 17:10
Sambo
1,2561427
As Hagen von Eitzen also pointed out, the problem lies in taking the following inequality:
$$
|a_i_1 + a_i_2 + ... + a_i_M-N| leq |s_M' - s_N| =
|a_N+1 + a_N+2 + ... + a_M'|
$$
The indices $i_1, ..., i_M-N$ are between $N+1$ and $M'$, and so the inequality would hold if all the $a_k$ were positive. However, this is not the case, and so we cannot take the inequality. E.g., we have:
$$
|1+1| nleq |1+(-1)+1|
$$
answered Jul 22 at 17:10
Sambo
1,2561427
answered Jul 22 at 17:10
Sambo
1,2561427
answered Jul 22 at 17:10
Sambo
1,2561427
answered Jul 22 at 17:10
Sambo
1,2561427
1,2561427
Thanks @Sambo .
– student
Jul 22 at 18:51
add a comment |Â
Thanks @Sambo .
– student
Jul 22 at 18:51
Thanks @Sambo .
– student
Jul 22 at 18:51
Thanks @Sambo .
– student
Jul 22 at 18:51
add a comment |Â
up vote
2
down vote
Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_i_1+dots|$ should be $|b_i_1+dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
add a comment |Â
up vote
2
down vote
Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_i_1+dots|$ should be $|b_i_1+dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_i_1+dots|$ should be $|b_i_1+dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
Hard to say exactly what's wrong because it's hard to know exactly what you had in mind - in particular it seems to me that the $|a_i_1+dots|$ should be $|b_i_1+dots|$. But if I look at what you wrote and try to make a correct proof out of it, here's a hint regarding where the error is: If $a_1=1$, $a_2=-1$ and $a_3=1$ then $$|a_1+a_3|>|a_1+a_2+a_3|.$$
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
answered Jul 22 at 17:12
David C. Ullrich
54.1k33481
54.1k33481
add a comment |Â
add a comment |Â
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3
You cannot conclude $ldots le |s_M'-s_N|$ because some of the non-occurring terms might have opposite sign to the occurring terms
– Hagen von Eitzen
Jul 22 at 17:05
Gotcha! Thanks.
– student
Jul 22 at 18:51