If $sum|a_n|$ diverges and $sum a_n$ converges then $sum a_n^+$ diverges












0












$begingroup$


The problem goes as follows:



Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.



Please do check the proof and make necessary corrections.



Proof:



We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.



Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$



We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .






sequences-and-series proof-verification







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  • 1




    $begingroup$
    More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
    $endgroup$
    – Did
    Dec 2 '18 at 9:17


















0












$begingroup$


The problem goes as follows:



Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.



Please do check the proof and make necessary corrections.



Proof:



We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.



Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$



We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .






sequences-and-series proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
    $endgroup$
    – Did
    Dec 2 '18 at 9:17
















0












0








0





$begingroup$


The problem goes as follows:



Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.



Please do check the proof and make necessary corrections.



Proof:



We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.



Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$



We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .






sequences-and-series proof-verification







share|cite|improve this question











$endgroup$




The problem goes as follows:



Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.



Please do check the proof and make necessary corrections.



Proof:



We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.



Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$



We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .






sequences-and-series proof-verification




sequences-and-series proof-verification






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edited Dec 2 '18 at 9:15









Did

247k23223459




247k23223459










asked Dec 2 '18 at 8:57









Subhasis BiswasSubhasis Biswas

488311




488311








  • 1




    $begingroup$
    More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
    $endgroup$
    – Did
    Dec 2 '18 at 9:17
















  • 1




    $begingroup$
    More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
    $endgroup$
    – Did
    Dec 2 '18 at 9:17










1




1




$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17






$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17












2 Answers
2






active

oldest

votes


















2












$begingroup$

For all $n in mathbb{N}$,



$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$



Therefore, for $N in mathbb{N}$ :



$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$



where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.






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  • 3




    $begingroup$
    Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
    $endgroup$
    – Martin R
    Dec 2 '18 at 9:18












  • $begingroup$
    can you please check the validity of the solution
    $endgroup$
    – Subhasis Biswas
    Dec 2 '18 at 9:35



















1












$begingroup$

(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For all $n in mathbb{N}$,



    $$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$



    Therefore, for $N in mathbb{N}$ :



    $$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$



    where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
      $endgroup$
      – Martin R
      Dec 2 '18 at 9:18












    • $begingroup$
      can you please check the validity of the solution
      $endgroup$
      – Subhasis Biswas
      Dec 2 '18 at 9:35
















    2












    $begingroup$

    For all $n in mathbb{N}$,



    $$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$



    Therefore, for $N in mathbb{N}$ :



    $$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$



    where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
      $endgroup$
      – Martin R
      Dec 2 '18 at 9:18












    • $begingroup$
      can you please check the validity of the solution
      $endgroup$
      – Subhasis Biswas
      Dec 2 '18 at 9:35














    2












    2








    2





    $begingroup$

    For all $n in mathbb{N}$,



    $$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$



    Therefore, for $N in mathbb{N}$ :



    $$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$



    where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.






    share|cite|improve this answer









    $endgroup$



    For all $n in mathbb{N}$,



    $$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$



    Therefore, for $N in mathbb{N}$ :



    $$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$



    where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 9:16









    jibounetjibounet

    5,6481032




    5,6481032








    • 3




      $begingroup$
      Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
      $endgroup$
      – Martin R
      Dec 2 '18 at 9:18












    • $begingroup$
      can you please check the validity of the solution
      $endgroup$
      – Subhasis Biswas
      Dec 2 '18 at 9:35














    • 3




      $begingroup$
      Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
      $endgroup$
      – Martin R
      Dec 2 '18 at 9:18












    • $begingroup$
      can you please check the validity of the solution
      $endgroup$
      – Subhasis Biswas
      Dec 2 '18 at 9:35








    3




    3




    $begingroup$
    Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
    $endgroup$
    – Martin R
    Dec 2 '18 at 9:18






    $begingroup$
    Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
    $endgroup$
    – Martin R
    Dec 2 '18 at 9:18














    $begingroup$
    can you please check the validity of the solution
    $endgroup$
    – Subhasis Biswas
    Dec 2 '18 at 9:35




    $begingroup$
    can you please check the validity of the solution
    $endgroup$
    – Subhasis Biswas
    Dec 2 '18 at 9:35











    1












    $begingroup$

    (1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      (1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        (1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.






        share|cite|improve this answer









        $endgroup$



        (1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 9:38









        modest_mildewmodest_mildew

        112




        112






























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