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






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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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  • 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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                                濃尾地震

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                                濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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