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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          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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