If $sum a_{n}$ is a series of positive terms and $b_{n} = (a_{1}+a_{2}+a_{3}+ dots +a_{n})/n$ then $sum...












1












$begingroup$


If $sum a_{n}$ is a series of positive terms and $b_{n} = (a_{1}+a_{2}+a_{3}+ dots +a_{n})/n$ then $sum b_{n}$ is divergent. (True/false)



The statement is given true with a proof in my book.
But I'm confused.
If I take the series $sum frac{1}{n^{2}}, $ for which sequence of partial sum is $a_{n} = (1)+(1/4)+(1/9)+(1/16)+ dots$ then,sequence of partial sum for $sum {b_{n}} $ will be $b_{n} = (1/2)+(1/8)+(1/18)+ dots$ which is convergent.
Then why is this statement true?






sequences-and-series







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    If $sum a_{n}$ is a series of positive terms and $b_{n} = (a_{1}+a_{2}+a_{3}+ dots +a_{n})/n$ then $sum b_{n}$ is divergent. (True/false)



    The statement is given true with a proof in my book.
    But I'm confused.
    If I take the series $sum frac{1}{n^{2}}, $ for which sequence of partial sum is $a_{n} = (1)+(1/4)+(1/9)+(1/16)+ dots$ then,sequence of partial sum for $sum {b_{n}} $ will be $b_{n} = (1/2)+(1/8)+(1/18)+ dots$ which is convergent.
    Then why is this statement true?






    sequences-and-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      If $sum a_{n}$ is a series of positive terms and $b_{n} = (a_{1}+a_{2}+a_{3}+ dots +a_{n})/n$ then $sum b_{n}$ is divergent. (True/false)



      The statement is given true with a proof in my book.
      But I'm confused.
      If I take the series $sum frac{1}{n^{2}}, $ for which sequence of partial sum is $a_{n} = (1)+(1/4)+(1/9)+(1/16)+ dots$ then,sequence of partial sum for $sum {b_{n}} $ will be $b_{n} = (1/2)+(1/8)+(1/18)+ dots$ which is convergent.
      Then why is this statement true?






      sequences-and-series







      share|cite|improve this question









      $endgroup$




      If $sum a_{n}$ is a series of positive terms and $b_{n} = (a_{1}+a_{2}+a_{3}+ dots +a_{n})/n$ then $sum b_{n}$ is divergent. (True/false)



      The statement is given true with a proof in my book.
      But I'm confused.
      If I take the series $sum frac{1}{n^{2}}, $ for which sequence of partial sum is $a_{n} = (1)+(1/4)+(1/9)+(1/16)+ dots$ then,sequence of partial sum for $sum {b_{n}} $ will be $b_{n} = (1/2)+(1/8)+(1/18)+ dots$ which is convergent.
      Then why is this statement true?






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 21:03









      MathsaddictMathsaddict

      3619




      3619






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          This follows by series comparison. Since $b_n ge frac{a_1}{n}$, the terms of the series $sum_n b_n$ are bounded from below by a multiple of the harmonic series, which is divergent,






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
            $endgroup$
            – John Omielan
            Jan 7 at 21:15





















          5












          $begingroup$

          Given that $a_i$ are all positive and



          $$b_n = left(a_1 + a_2 + a_3 + cdots + a_nright)/n tag{1}label{eq1}$$



          Then note that $b_n ge frac{a_1}{n}$ for all $n ge 1$. Thus,



          $$sum_{i = 1}^{n} b_i ge sum_{i = 1}^{n} frac{a_1}{n} = a_1 sum_{i = 1}^{n} frac{1}{n} tag{2}label{eq2}$$



          Note that the latter part is the harmonic series, which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, I have read this proof. But, what is wrong with my example?
            $endgroup$
            – Mathsaddict
            Jan 7 at 21:23










          • $begingroup$
            Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
            $endgroup$
            – John Omielan
            Jan 7 at 21:32








          • 1




            $begingroup$
            @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
            $endgroup$
            – John Omielan
            Jan 7 at 21:41








          • 1




            $begingroup$
            yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
            $endgroup$
            – Mathsaddict
            Jan 8 at 1:48










          • $begingroup$
            @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
            $endgroup$
            – John Omielan
            Jan 8 at 5:46













          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          This follows by series comparison. Since $b_n ge frac{a_1}{n}$, the terms of the series $sum_n b_n$ are bounded from below by a multiple of the harmonic series, which is divergent,






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
            $endgroup$
            – John Omielan
            Jan 7 at 21:15


















          5












          $begingroup$

          This follows by series comparison. Since $b_n ge frac{a_1}{n}$, the terms of the series $sum_n b_n$ are bounded from below by a multiple of the harmonic series, which is divergent,






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
            $endgroup$
            – John Omielan
            Jan 7 at 21:15
















          5












          5








          5





          $begingroup$

          This follows by series comparison. Since $b_n ge frac{a_1}{n}$, the terms of the series $sum_n b_n$ are bounded from below by a multiple of the harmonic series, which is divergent,






          share|cite|improve this answer









          $endgroup$



          This follows by series comparison. Since $b_n ge frac{a_1}{n}$, the terms of the series $sum_n b_n$ are bounded from below by a multiple of the harmonic series, which is divergent,







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 21:12









          Hans EnglerHans Engler

          10.5k11836




          10.5k11836












          • $begingroup$
            I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
            $endgroup$
            – John Omielan
            Jan 7 at 21:15




















          • $begingroup$
            I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
            $endgroup$
            – John Omielan
            Jan 7 at 21:15


















          $begingroup$
          I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
          $endgroup$
          – John Omielan
          Jan 7 at 21:15






          $begingroup$
          I was just adding my answer when you posted yours. As they're basically the same, although I provide a bit more detail, I can delete mine if you wish.
          $endgroup$
          – John Omielan
          Jan 7 at 21:15













          5












          $begingroup$

          Given that $a_i$ are all positive and



          $$b_n = left(a_1 + a_2 + a_3 + cdots + a_nright)/n tag{1}label{eq1}$$



          Then note that $b_n ge frac{a_1}{n}$ for all $n ge 1$. Thus,



          $$sum_{i = 1}^{n} b_i ge sum_{i = 1}^{n} frac{a_1}{n} = a_1 sum_{i = 1}^{n} frac{1}{n} tag{2}label{eq2}$$



          Note that the latter part is the harmonic series, which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, I have read this proof. But, what is wrong with my example?
            $endgroup$
            – Mathsaddict
            Jan 7 at 21:23










          • $begingroup$
            Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
            $endgroup$
            – John Omielan
            Jan 7 at 21:32








          • 1




            $begingroup$
            @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
            $endgroup$
            – John Omielan
            Jan 7 at 21:41








          • 1




            $begingroup$
            yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
            $endgroup$
            – Mathsaddict
            Jan 8 at 1:48










          • $begingroup$
            @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
            $endgroup$
            – John Omielan
            Jan 8 at 5:46


















          5












          $begingroup$

          Given that $a_i$ are all positive and



          $$b_n = left(a_1 + a_2 + a_3 + cdots + a_nright)/n tag{1}label{eq1}$$



          Then note that $b_n ge frac{a_1}{n}$ for all $n ge 1$. Thus,



          $$sum_{i = 1}^{n} b_i ge sum_{i = 1}^{n} frac{a_1}{n} = a_1 sum_{i = 1}^{n} frac{1}{n} tag{2}label{eq2}$$



          Note that the latter part is the harmonic series, which diverges.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, I have read this proof. But, what is wrong with my example?
            $endgroup$
            – Mathsaddict
            Jan 7 at 21:23










          • $begingroup$
            Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
            $endgroup$
            – John Omielan
            Jan 7 at 21:32








          • 1




            $begingroup$
            @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
            $endgroup$
            – John Omielan
            Jan 7 at 21:41








          • 1




            $begingroup$
            yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
            $endgroup$
            – Mathsaddict
            Jan 8 at 1:48










          • $begingroup$
            @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
            $endgroup$
            – John Omielan
            Jan 8 at 5:46
















          5












          5








          5





          $begingroup$

          Given that $a_i$ are all positive and



          $$b_n = left(a_1 + a_2 + a_3 + cdots + a_nright)/n tag{1}label{eq1}$$



          Then note that $b_n ge frac{a_1}{n}$ for all $n ge 1$. Thus,



          $$sum_{i = 1}^{n} b_i ge sum_{i = 1}^{n} frac{a_1}{n} = a_1 sum_{i = 1}^{n} frac{1}{n} tag{2}label{eq2}$$



          Note that the latter part is the harmonic series, which diverges.






          share|cite|improve this answer









          $endgroup$



          Given that $a_i$ are all positive and



          $$b_n = left(a_1 + a_2 + a_3 + cdots + a_nright)/n tag{1}label{eq1}$$



          Then note that $b_n ge frac{a_1}{n}$ for all $n ge 1$. Thus,



          $$sum_{i = 1}^{n} b_i ge sum_{i = 1}^{n} frac{a_1}{n} = a_1 sum_{i = 1}^{n} frac{1}{n} tag{2}label{eq2}$$



          Note that the latter part is the harmonic series, which diverges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 21:13









          John OmielanJohn Omielan

          2,779212




          2,779212












          • $begingroup$
            Yes, I have read this proof. But, what is wrong with my example?
            $endgroup$
            – Mathsaddict
            Jan 7 at 21:23










          • $begingroup$
            Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
            $endgroup$
            – John Omielan
            Jan 7 at 21:32








          • 1




            $begingroup$
            @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
            $endgroup$
            – John Omielan
            Jan 7 at 21:41








          • 1




            $begingroup$
            yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
            $endgroup$
            – Mathsaddict
            Jan 8 at 1:48










          • $begingroup$
            @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
            $endgroup$
            – John Omielan
            Jan 8 at 5:46




















          • $begingroup$
            Yes, I have read this proof. But, what is wrong with my example?
            $endgroup$
            – Mathsaddict
            Jan 7 at 21:23










          • $begingroup$
            Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
            $endgroup$
            – John Omielan
            Jan 7 at 21:32








          • 1




            $begingroup$
            @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
            $endgroup$
            – John Omielan
            Jan 7 at 21:41








          • 1




            $begingroup$
            yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
            $endgroup$
            – Mathsaddict
            Jan 8 at 1:48










          • $begingroup$
            @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
            $endgroup$
            – John Omielan
            Jan 8 at 5:46


















          $begingroup$
          Yes, I have read this proof. But, what is wrong with my example?
          $endgroup$
          – Mathsaddict
          Jan 7 at 21:23




          $begingroup$
          Yes, I have read this proof. But, what is wrong with my example?
          $endgroup$
          – Mathsaddict
          Jan 7 at 21:23












          $begingroup$
          Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
          $endgroup$
          – John Omielan
          Jan 7 at 21:32






          $begingroup$
          Note that $sum frac{1}{n}$ diverges, but $sum frac{1}{n^2}$ converges, so you can select $a_i$ for the latter case which will allow $sum b_i$ to converge.
          $endgroup$
          – John Omielan
          Jan 7 at 21:32






          1




          1




          $begingroup$
          @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
          $endgroup$
          – John Omielan
          Jan 7 at 21:41






          $begingroup$
          @Mathsaddict I misread your original text. Using $a_i = frac{1}{i^2}$, then $b_1 = frac{a_1}{1} = 1$, not $frac{1}{2}$. Next, $b_2 = frac{a_1 + a_2}{2} = frac{1 + 1/4}{2} = frac{5}{8}$, not $frac{1}{8}$. If you continue, you will see the next values are larger than you indicated, and don't become small enough, fast enough, for the sum to converge, as another answer here already states.
          $endgroup$
          – John Omielan
          Jan 7 at 21:41






          1




          1




          $begingroup$
          yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
          $endgroup$
          – Mathsaddict
          Jan 8 at 1:48




          $begingroup$
          yes, I calculated $b_{n}$ incorrectly. So, this series $sum b_{n}$ will always be divergent. It doesn't matter if $sum a_{n}$ is convergent or not. Right?
          $endgroup$
          – Mathsaddict
          Jan 8 at 1:48












          $begingroup$
          @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
          $endgroup$
          – John Omielan
          Jan 8 at 5:46






          $begingroup$
          @Mathsaddict Yes, you are correct. That is what my proof, and basically the same one of Hans Engler, shows.
          $endgroup$
          – John Omielan
          Jan 8 at 5:46




















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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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