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.







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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














up vote
0
down vote

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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.







share|cite|improve this question















  • 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












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.







share|cite|improve this question











$sum_n=1^infty a_n$ 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.









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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












  • 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










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:



$$
|1+1| nleq |1+(-1)+1|
$$






share|cite|improve this answer





















  • Thanks @Sambo .
    – student
    Jul 22 at 18:51

















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 $$|a_1+a_3|>|a_1+a_2+a_3|.$$






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    2 Answers
    2






    active

    oldest

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    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:



    $$
    |1+1| nleq |1+(-1)+1|
    $$






    share|cite|improve this answer





















    • Thanks @Sambo .
      – student
      Jul 22 at 18:51














    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:



    $$
    |1+1| nleq |1+(-1)+1|
    $$






    share|cite|improve this answer





















    • Thanks @Sambo .
      – student
      Jul 22 at 18:51












    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:



    $$
    |1+1| nleq |1+(-1)+1|
    $$






    share|cite|improve this answer













    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:



    $$
    |1+1| nleq |1+(-1)+1|
    $$







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 22 at 17:10









    Sambo

    1,2561427




    1,2561427











    • Thanks @Sambo .
      – student
      Jul 22 at 18:51
















    • Thanks @Sambo .
      – student
      Jul 22 at 18:51















    Thanks @Sambo .
    – student
    Jul 22 at 18:51




    Thanks @Sambo .
    – student
    Jul 22 at 18:51










    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 $$|a_1+a_3|>|a_1+a_2+a_3|.$$






    share|cite|improve this answer

























      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 $$|a_1+a_3|>|a_1+a_2+a_3|.$$






      share|cite|improve this answer























        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 $$|a_1+a_3|>|a_1+a_2+a_3|.$$






        share|cite|improve this answer













        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 $$|a_1+a_3|>|a_1+a_2+a_3|.$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 22 at 17:12









        David C. Ullrich

        54.1k33481




        54.1k33481






















             

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