Attempt on modified harmonic series












2















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




Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}



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



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

Thanks in advance.






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







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  • 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
    26 mins ago












  • @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
    20 mins ago
















2















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




Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}



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



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

Thanks in advance.






calculus sequences-and-series proof-verification 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.
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  • 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
    26 mins ago












  • @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
    20 mins ago














2












2








2








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




Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}



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



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

Thanks in advance.






calculus sequences-and-series proof-verification 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 $left(1+frac{1}{2}-frac{1}{3} right) + left(frac{1}{4}+frac{1}{5}-frac{1}{6} right)+left(frac{1}{7}+frac{1}{8}-frac{1}{9} right)+ldots quad$ converge?




Here's my attempt at a solution: begin{eqnarray*}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)+ldots &= 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} end{eqnarray*}



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



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

Thanks in advance.






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




calculus sequences-and-series proof-verification 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.











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asked 58 mins ago









Raúl Astete

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235




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New contributor





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












  • 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
    26 mins ago












  • @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
    20 mins ago


















  • 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
    26 mins ago












  • @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
    20 mins ago
















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
26 mins ago






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
26 mins ago














@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
20 mins ago




@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
20 mins ago










3 Answers
3






active

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3














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














    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





























      1














      Yes, your answer is correct.



      By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



      Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



      Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






      share|cite|improve this answer





















        Your Answer





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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

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        active

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        active

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        3














        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




























          3














          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


























            3












            3








            3






            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 31 mins ago

























            answered 38 mins ago









            Ben W

            1,800514




            1,800514























                2














                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


























                  2














                  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
























                    2












                    2








                    2






                    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 28 mins ago









                    xbh

                    5,6951522




                    5,6951522























                        1














                        Yes, your answer is correct.



                        By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                        Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                        Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                        share|cite|improve this answer


























                          1














                          Yes, your answer is correct.



                          By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                          Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                          Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Yes, your answer is correct.



                            By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                            Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                            Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.






                            share|cite|improve this answer












                            Yes, your answer is correct.



                            By using Cauchy's Convergence test, for $sum_{n=1}^inftydfrac1n$



                            Taking $$S_{2n}-S_n=sum_{n=1}^{2n}dfrac1n-sum_{n=1}^ndfrac1n=dfrac{1}{n+1}+dfrac{1}{n+2}+....+dfrac{1}{2n}gedfrac{1}{2n}+....+dfrac{1}{2n}=dfrac{n}{2n}=dfrac{1}{2}$$



                            Since, there is no number $N$ that satisfies Cauchy's condition, the series diverges.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 42 mins ago









                            Key Flex

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