$sum a_{n}$ converges but $sum a_{n}^2 $ diverges?











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I have to give an example of a convergent series $sum a_{n}$ for which $sum a_{n}^2 $ diverges.



I think that such a series cannot exist because if $sum a_{n}$ converges absolutely then $sum a_{n}^2 $ will always converge right?










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  • 1




    The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
    – Saucy O'Path
    Nov 4 at 23:05












  • @max_zorn That's if $sum a_n$ converges absolutely.
    – Robert Israel
    Nov 4 at 23:09










  • Yes, you are right Robert, sorry!
    – max_zorn
    Nov 4 at 23:12















up vote
2
down vote

favorite












I have to give an example of a convergent series $sum a_{n}$ for which $sum a_{n}^2 $ diverges.



I think that such a series cannot exist because if $sum a_{n}$ converges absolutely then $sum a_{n}^2 $ will always converge right?










share|cite|improve this question









New contributor




Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
    – Saucy O'Path
    Nov 4 at 23:05












  • @max_zorn That's if $sum a_n$ converges absolutely.
    – Robert Israel
    Nov 4 at 23:09










  • Yes, you are right Robert, sorry!
    – max_zorn
    Nov 4 at 23:12













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have to give an example of a convergent series $sum a_{n}$ for which $sum a_{n}^2 $ diverges.



I think that such a series cannot exist because if $sum a_{n}$ converges absolutely then $sum a_{n}^2 $ will always converge right?










share|cite|improve this question









New contributor




Snop D. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have to give an example of a convergent series $sum a_{n}$ for which $sum a_{n}^2 $ diverges.



I think that such a series cannot exist because if $sum a_{n}$ converges absolutely then $sum a_{n}^2 $ will always converge right?







real-analysis sequences-and-series limits convergence






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edited Nov 5 at 7:59









user21820

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asked Nov 4 at 23:03









Snop D.

111




111




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  • 1




    The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
    – Saucy O'Path
    Nov 4 at 23:05












  • @max_zorn That's if $sum a_n$ converges absolutely.
    – Robert Israel
    Nov 4 at 23:09










  • Yes, you are right Robert, sorry!
    – max_zorn
    Nov 4 at 23:12














  • 1




    The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
    – Saucy O'Path
    Nov 4 at 23:05












  • @max_zorn That's if $sum a_n$ converges absolutely.
    – Robert Israel
    Nov 4 at 23:09










  • Yes, you are right Robert, sorry!
    – max_zorn
    Nov 4 at 23:12








1




1




The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
Nov 4 at 23:05






The implication you said after "because" is true but it's still possible that $sum_n a_n$ converges but not absolutely and $sum_n a_n^2$ does not converge (absolutely nor by any means).
– Saucy O'Path
Nov 4 at 23:05














@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09




@max_zorn That's if $sum a_n$ converges absolutely.
– Robert Israel
Nov 4 at 23:09












Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12




Yes, you are right Robert, sorry!
– max_zorn
Nov 4 at 23:12










4 Answers
4






active

oldest

votes

















up vote
12
down vote













The alternating series test gives a wealth of examples. Take
$$
a_n=(-1)^n/sqrt{n}
$$

for example.






share|cite|improve this answer





















  • You beat me to it by $14$ seconds.
    – Robert Israel
    Nov 4 at 23:07






  • 5




    Are there any that don't involve a $(-1)^n$ term?
    – Ingolifs
    Nov 5 at 4:27






  • 1




    @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
    – user21820
    Nov 5 at 8:00


















up vote
10
down vote













Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$






share|cite|improve this answer




























    up vote
    3
    down vote













    More generally,
    if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
    then
    $sum_{n=1}^{infty} a_n
    $

    converges for integer $m ge 0$
    and
    $sum_{n=1}^{infty} a_n^{2m}
    $

    diverges.






    share|cite|improve this answer




























      up vote
      0
      down vote













      The simplest example I have in mind is :
      $$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$



      Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
      Indeed, if $sum_{n=1}^{+infty} a_n $ is ACV then, $sum_{n=1}^{+infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.



      You can also ask yourself a more general question :
      What are the function $f:mathbb R rightarrow mathbb R$ such that for all
      $sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).



      It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,



      $$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$



      $$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$



      $$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$



      $$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$






      share|cite|improve this answer










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      • 1




        $sum_n a_n^2$ converges for this one.
        – Robert Israel
        Nov 4 at 23:07










      • oups I answer too fast ^^
        – Can I play with Mathness
        Nov 4 at 23:10






      • 3




        I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
        – Matt Samuel
        Nov 5 at 1:23






      • 2




        While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
        – eyeballfrog
        Nov 5 at 2:33






      • 1




        I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
        – Winther
        Nov 5 at 2:34













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      12
      down vote













      The alternating series test gives a wealth of examples. Take
      $$
      a_n=(-1)^n/sqrt{n}
      $$

      for example.






      share|cite|improve this answer





















      • You beat me to it by $14$ seconds.
        – Robert Israel
        Nov 4 at 23:07






      • 5




        Are there any that don't involve a $(-1)^n$ term?
        – Ingolifs
        Nov 5 at 4:27






      • 1




        @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
        – user21820
        Nov 5 at 8:00















      up vote
      12
      down vote













      The alternating series test gives a wealth of examples. Take
      $$
      a_n=(-1)^n/sqrt{n}
      $$

      for example.






      share|cite|improve this answer





















      • You beat me to it by $14$ seconds.
        – Robert Israel
        Nov 4 at 23:07






      • 5




        Are there any that don't involve a $(-1)^n$ term?
        – Ingolifs
        Nov 5 at 4:27






      • 1




        @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
        – user21820
        Nov 5 at 8:00













      up vote
      12
      down vote










      up vote
      12
      down vote









      The alternating series test gives a wealth of examples. Take
      $$
      a_n=(-1)^n/sqrt{n}
      $$

      for example.






      share|cite|improve this answer












      The alternating series test gives a wealth of examples. Take
      $$
      a_n=(-1)^n/sqrt{n}
      $$

      for example.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 4 at 23:05









      Foobaz John

      19.3k41248




      19.3k41248












      • You beat me to it by $14$ seconds.
        – Robert Israel
        Nov 4 at 23:07






      • 5




        Are there any that don't involve a $(-1)^n$ term?
        – Ingolifs
        Nov 5 at 4:27






      • 1




        @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
        – user21820
        Nov 5 at 8:00


















      • You beat me to it by $14$ seconds.
        – Robert Israel
        Nov 4 at 23:07






      • 5




        Are there any that don't involve a $(-1)^n$ term?
        – Ingolifs
        Nov 5 at 4:27






      • 1




        @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
        – user21820
        Nov 5 at 8:00
















      You beat me to it by $14$ seconds.
      – Robert Israel
      Nov 4 at 23:07




      You beat me to it by $14$ seconds.
      – Robert Israel
      Nov 4 at 23:07




      5




      5




      Are there any that don't involve a $(-1)^n$ term?
      – Ingolifs
      Nov 5 at 4:27




      Are there any that don't involve a $(-1)^n$ term?
      – Ingolifs
      Nov 5 at 4:27




      1




      1




      @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
      – user21820
      Nov 5 at 8:00




      @Ingolifs: $sin$. But if the sequence is non-negative, it is a basic exercise to prove that the implication holds.
      – user21820
      Nov 5 at 8:00










      up vote
      10
      down vote













      Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$






      share|cite|improve this answer

























        up vote
        10
        down vote













        Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$






        share|cite|improve this answer























          up vote
          10
          down vote










          up vote
          10
          down vote









          Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$






          share|cite|improve this answer












          Try $$sum_{n=1}^infty frac{(-1)^n}{sqrt{n}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 4 at 23:06









          Robert Israel

          312k23206452




          312k23206452






















              up vote
              3
              down vote













              More generally,
              if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
              then
              $sum_{n=1}^{infty} a_n
              $

              converges for integer $m ge 0$
              and
              $sum_{n=1}^{infty} a_n^{2m}
              $

              diverges.






              share|cite|improve this answer

























                up vote
                3
                down vote













                More generally,
                if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
                then
                $sum_{n=1}^{infty} a_n
                $

                converges for integer $m ge 0$
                and
                $sum_{n=1}^{infty} a_n^{2m}
                $

                diverges.






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  More generally,
                  if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
                  then
                  $sum_{n=1}^{infty} a_n
                  $

                  converges for integer $m ge 0$
                  and
                  $sum_{n=1}^{infty} a_n^{2m}
                  $

                  diverges.






                  share|cite|improve this answer












                  More generally,
                  if $a_n = dfrac{(-1)^n}{n^{1/(2m)}}$,
                  then
                  $sum_{n=1}^{infty} a_n
                  $

                  converges for integer $m ge 0$
                  and
                  $sum_{n=1}^{infty} a_n^{2m}
                  $

                  diverges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 4 at 23:29









                  marty cohen

                  71k546122




                  71k546122






















                      up vote
                      0
                      down vote













                      The simplest example I have in mind is :
                      $$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$



                      Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
                      Indeed, if $sum_{n=1}^{+infty} a_n $ is ACV then, $sum_{n=1}^{+infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.



                      You can also ask yourself a more general question :
                      What are the function $f:mathbb R rightarrow mathbb R$ such that for all
                      $sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).



                      It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$



                      $$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$



                      $$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$






                      share|cite|improve this answer










                      New contributor




                      Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.














                      • 1




                        $sum_n a_n^2$ converges for this one.
                        – Robert Israel
                        Nov 4 at 23:07










                      • oups I answer too fast ^^
                        – Can I play with Mathness
                        Nov 4 at 23:10






                      • 3




                        I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                        – Matt Samuel
                        Nov 5 at 1:23






                      • 2




                        While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                        – eyeballfrog
                        Nov 5 at 2:33






                      • 1




                        I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                        – Winther
                        Nov 5 at 2:34

















                      up vote
                      0
                      down vote













                      The simplest example I have in mind is :
                      $$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$



                      Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
                      Indeed, if $sum_{n=1}^{+infty} a_n $ is ACV then, $sum_{n=1}^{+infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.



                      You can also ask yourself a more general question :
                      What are the function $f:mathbb R rightarrow mathbb R$ such that for all
                      $sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).



                      It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$



                      $$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$



                      $$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$






                      share|cite|improve this answer










                      New contributor




                      Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.














                      • 1




                        $sum_n a_n^2$ converges for this one.
                        – Robert Israel
                        Nov 4 at 23:07










                      • oups I answer too fast ^^
                        – Can I play with Mathness
                        Nov 4 at 23:10






                      • 3




                        I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                        – Matt Samuel
                        Nov 5 at 1:23






                      • 2




                        While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                        – eyeballfrog
                        Nov 5 at 2:33






                      • 1




                        I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                        – Winther
                        Nov 5 at 2:34















                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      The simplest example I have in mind is :
                      $$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$



                      Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
                      Indeed, if $sum_{n=1}^{+infty} a_n $ is ACV then, $sum_{n=1}^{+infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.



                      You can also ask yourself a more general question :
                      What are the function $f:mathbb R rightarrow mathbb R$ such that for all
                      $sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).



                      It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$



                      $$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$



                      $$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$






                      share|cite|improve this answer










                      New contributor




                      Can I play with Mathness is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      The simplest example I have in mind is :
                      $$sum_{n=1}^{+infty}frac{(-1)^n}{sqrt{n}} $$



                      Beware that convergent (CV) and absolutely convergent (ACV) can be very different.
                      Indeed, if $sum_{n=1}^{+infty} a_n $ is ACV then, $sum_{n=1}^{+infty} a_n^2$ is also ACV. To prove it, you can simply notice that, since $lim a_n=0$ then for $n$ big enough, $|a_n^2|<|a_n|$.



                      You can also ask yourself a more general question :
                      What are the function $f:mathbb R rightarrow mathbb R$ such that for all
                      $sum a_n$ CV (resp. ACV), $sum f(a_n)$ is CV (resp. ACV).



                      It is a (difficult) exercise to show that for $f:mathbb Rrightarrow mathbb R$,



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~CV quad text{iff} quad exists eta>0,exists lambdain mathbb R,forall xin ]-eta,eta[, quad f(x)=lambda x $$



                      $$sum a_n ~CV Rightarrow sum f(a_n) ~ACV quad text{iff} quad exists eta>0,forall xin ]-eta,eta[, quad f(x)=0 quadquadquad~~$$



                      $$sum a_n ~ACV Rightarrow sum f(a_n) ~ACV quad text{iff} quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad$$ $$quad f(0)=0 text{ and }exists eta>0,exists M >0,forall xin ]-eta,eta[, |f(x)|leq M |x| $$



                      $$sum a_n ACV Rightarrow sum f(a_n) ~CV quad text{iff} quad sum a_n ~ACV Rightarrow sum f(a_n) ~ACV~~~~~~~~~~~~~~ $$







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                      edited Nov 4 at 23:20





















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                      answered Nov 4 at 23:05









                      Can I play with Mathness

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                      • 1




                        $sum_n a_n^2$ converges for this one.
                        – Robert Israel
                        Nov 4 at 23:07










                      • oups I answer too fast ^^
                        – Can I play with Mathness
                        Nov 4 at 23:10






                      • 3




                        I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                        – Matt Samuel
                        Nov 5 at 1:23






                      • 2




                        While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                        – eyeballfrog
                        Nov 5 at 2:33






                      • 1




                        I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                        – Winther
                        Nov 5 at 2:34
















                      • 1




                        $sum_n a_n^2$ converges for this one.
                        – Robert Israel
                        Nov 4 at 23:07










                      • oups I answer too fast ^^
                        – Can I play with Mathness
                        Nov 4 at 23:10






                      • 3




                        I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                        – Matt Samuel
                        Nov 5 at 1:23






                      • 2




                        While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                        – eyeballfrog
                        Nov 5 at 2:33






                      • 1




                        I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                        – Winther
                        Nov 5 at 2:34










                      1




                      1




                      $sum_n a_n^2$ converges for this one.
                      – Robert Israel
                      Nov 4 at 23:07




                      $sum_n a_n^2$ converges for this one.
                      – Robert Israel
                      Nov 4 at 23:07












                      oups I answer too fast ^^
                      – Can I play with Mathness
                      Nov 4 at 23:10




                      oups I answer too fast ^^
                      – Can I play with Mathness
                      Nov 4 at 23:10




                      3




                      3




                      I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                      – Matt Samuel
                      Nov 5 at 1:23




                      I have a PhD in math and my eyes glaze over when reading this post because of the unnecessary jumble of symbols. Words are generally preferable for longer statements.
                      – Matt Samuel
                      Nov 5 at 1:23




                      2




                      2




                      While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                      – eyeballfrog
                      Nov 5 at 2:33




                      While I realize it's a standard notation, $xin]-eta,eta[$ just looks so horrible--the unbalanced brackets make it hard to parse. $xin(-eta,eta)$ is much better, though $|x| < eta$ is better than both.
                      – eyeballfrog
                      Nov 5 at 2:33




                      1




                      1




                      I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                      – Winther
                      Nov 5 at 2:34






                      I'm in agreement with the comment by Matt above, for me it's almost unreadable (in the sense that it takes way too much effort to parse it). A (IMO) better way of presenting the first statement would be something like "If $sum a_n$ converges then $sum f(a_n)$ converges if $f(x)$ is proportional to $x$ in a neighborhood of $x=0$."
                      – Winther
                      Nov 5 at 2:34












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