Does $(1+frac12-frac13) + (frac14+frac15-frac16)+(frac17+frac18-frac19)+cdots$ converge?












5















Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?




Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.



Is this right? Which other convergence tests could be used?






calculus sequences-and-series proof-verification convergence divergent-series







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  • 2




    your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    yesterday












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    yesterday






  • 4




    @Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
    – AlephNull
    22 hours ago
















5















Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?




Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.



Is this right? Which other convergence tests could be used?






calculus sequences-and-series proof-verification convergence divergent-series







share|cite|improve this question









New contributor




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
















  • 2




    your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    yesterday












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    yesterday






  • 4




    @Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
    – AlephNull
    22 hours ago














5












5








5


1






Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?




Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.



Is this right? Which other convergence tests could be used?






calculus sequences-and-series proof-verification convergence divergent-series







share|cite|improve this question









New contributor




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












Does the series $$S=left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9}right)+cdots$$ converge?




Here's my attempt at a solution: $$S = sum_{n=1}^{infty}frac{1}{n}-2sum_{n=1}^{infty}frac{1}{3n}=sum_{n=1}^{infty}frac{1}{3n}=frac{1}{3}sum_{n=1}^{infty}frac{1}{n}$$



As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$.



Is this right? Which other convergence tests could be used?






calculus sequences-and-series proof-verification convergence divergent-series




calculus sequences-and-series proof-verification convergence divergent-series






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




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









share|cite|improve this question




share|cite|improve this question








edited 16 hours ago









Did

246k23221455




246k23221455






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asked yesterday









Raúl Astete

496




496




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New contributor





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






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Check out our Code of Conduct.








  • 2




    your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    yesterday












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    yesterday






  • 4




    @Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
    – AlephNull
    22 hours ago














  • 2




    your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
    – Masacroso
    yesterday












  • @Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
    – Saad
    yesterday






  • 4




    @Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
    – AlephNull
    22 hours ago








2




2




your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday






your solution is not correct. You cannot change the order of summation of a series that doesn't converge absolutely, neither make (standard) arithmetic with divergent series
– Masacroso
yesterday














@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday




@Masacroso Is it not fine to do something along the lines of " Suppose the following series converges. Then in particular, it converges absolutely, so we can reorder the series in any way. One particular ordering gives a scalar multiple of the harmonic series which diverges, so the given series cannot converge".
– Saad
yesterday




4




4




@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago




@Saad That is not a valid argument because convergence doesn't imply absolute convergence (you've got it the wrong way round).
– AlephNull
22 hours ago










2 Answers
2






active

oldest

votes


















11














Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
$$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
Simplify what is in the parentheses and then evaluate in the usual way.



Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






share|cite|improve this answer



















  • 2




    In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
    – zwim
    yesterday










  • @zwim quite true, that is a nice little shortcut
    – Ben W
    20 hours ago










  • Simplify and it behaves like $1/n$, so it diverges.
    – ncmathsadist
    16 hours ago



















4














The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
$$
S_n = H_{3n} - frac 23 H_n,
$$
then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
$$
S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
$$






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

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    active

    oldest

    votes









    11














    Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
    $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
    Simplify what is in the parentheses and then evaluate in the usual way.



    Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






    share|cite|improve this answer



















    • 2




      In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
      – zwim
      yesterday










    • @zwim quite true, that is a nice little shortcut
      – Ben W
      20 hours ago










    • Simplify and it behaves like $1/n$, so it diverges.
      – ncmathsadist
      16 hours ago
















    11














    Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
    $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
    Simplify what is in the parentheses and then evaluate in the usual way.



    Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






    share|cite|improve this answer



















    • 2




      In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
      – zwim
      yesterday










    • @zwim quite true, that is a nice little shortcut
      – Ben W
      20 hours ago










    • Simplify and it behaves like $1/n$, so it diverges.
      – ncmathsadist
      16 hours ago














    11












    11








    11






    Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
    $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
    Simplify what is in the parentheses and then evaluate in the usual way.



    Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.






    share|cite|improve this answer














    Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows:
    $$sum_{n=0}^inftyleft(frac{1}{3n+1}+frac{1}{3n+2}-frac{1}{3n+3}right)$$
    Simplify what is in the parentheses and then evaluate in the usual way.



    Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered yesterday









    Ben W

    1,995615




    1,995615








    • 2




      In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
      – zwim
      yesterday










    • @zwim quite true, that is a nice little shortcut
      – Ben W
      20 hours ago










    • Simplify and it behaves like $1/n$, so it diverges.
      – ncmathsadist
      16 hours ago














    • 2




      In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
      – zwim
      yesterday










    • @zwim quite true, that is a nice little shortcut
      – Ben W
      20 hours ago










    • Simplify and it behaves like $1/n$, so it diverges.
      – ncmathsadist
      16 hours ago








    2




    2




    In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
    – zwim
    yesterday




    In this case you can simply say $frac 1{3n+2}-frac 1{3n+3}>0$ thus the series is $>sumfrac 1{3n+1}$ which is a divergent series. I mean we do not even need to calculate that the parenthesis is equivalent to $frac {9n^2}{27n^3}$.
    – zwim
    yesterday












    @zwim quite true, that is a nice little shortcut
    – Ben W
    20 hours ago




    @zwim quite true, that is a nice little shortcut
    – Ben W
    20 hours ago












    Simplify and it behaves like $1/n$, so it diverges.
    – ncmathsadist
    16 hours ago




    Simplify and it behaves like $1/n$, so it diverges.
    – ncmathsadist
    16 hours ago











    4














    The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



    Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
    $$
    S_n = H_{3n} - frac 23 H_n,
    $$
    then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
    $$
    S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
    $$






    share|cite|improve this answer


























      4














      The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



      Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
      $$
      S_n = H_{3n} - frac 23 H_n,
      $$
      then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
      $$
      S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
      $$






      share|cite|improve this answer
























        4












        4








        4






        The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



        Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
        $$
        S_n = H_{3n} - frac 23 H_n,
        $$
        then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
        $$
        S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
        $$






        share|cite|improve this answer












        The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series.



        Let $H_n = sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the
        $$
        S_n = H_{3n} - frac 23 H_n,
        $$
        then use the asymptotic expression $H_n = log n + gamma + varepsilon_n$ where $varepsilon_n to 0 [n to infty]$, we have
        $$
        S_n = log (3n) -frac 23log n + frac 13 gamma + varepsilon_{3n} - frac 23 varepsilon_n = log(3n^{1/3}) + frac 13 gamma + alpha_n xrightarrow{n to infty} +infty.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        xbh

        5,7651522




        5,7651522






















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