Divergent series whose terms converge to zero












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I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?






calculus sequences-and-series convergence examples-counterexamples







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  • 1




    $begingroup$
    What tests have you done? The integral test is usually done pretty early in this sequence.
    $endgroup$
    – rogerl
    Jul 14 '15 at 1:31










  • $begingroup$
    What may be simpler than $1/n$?
    $endgroup$
    – Michael Galuza
    Jul 14 '15 at 1:31










  • $begingroup$
    You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
    $endgroup$
    – Alex R.
    Jul 14 '15 at 1:33










  • $begingroup$
    Good points. I guess there really is no reason not to use $sum 1/n$.
    $endgroup$
    – TorsionSquid
    Jul 14 '15 at 1:40










  • $begingroup$
    In any event: have you seen this?
    $endgroup$
    – J. M. is not a mathematician
    Jul 14 '15 at 2:51
















0












$begingroup$


I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?






calculus sequences-and-series convergence examples-counterexamples







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What tests have you done? The integral test is usually done pretty early in this sequence.
    $endgroup$
    – rogerl
    Jul 14 '15 at 1:31










  • $begingroup$
    What may be simpler than $1/n$?
    $endgroup$
    – Michael Galuza
    Jul 14 '15 at 1:31










  • $begingroup$
    You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
    $endgroup$
    – Alex R.
    Jul 14 '15 at 1:33










  • $begingroup$
    Good points. I guess there really is no reason not to use $sum 1/n$.
    $endgroup$
    – TorsionSquid
    Jul 14 '15 at 1:40










  • $begingroup$
    In any event: have you seen this?
    $endgroup$
    – J. M. is not a mathematician
    Jul 14 '15 at 2:51














0












0








0





$begingroup$


I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?






calculus sequences-and-series convergence examples-counterexamples







share|cite|improve this question











$endgroup$




I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?






calculus sequences-and-series convergence examples-counterexamples




calculus sequences-and-series convergence examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 9:48









Martin Sleziak

44.8k10119272




44.8k10119272










asked Jul 14 '15 at 1:28









TorsionSquidTorsionSquid

2,370721




2,370721








  • 1




    $begingroup$
    What tests have you done? The integral test is usually done pretty early in this sequence.
    $endgroup$
    – rogerl
    Jul 14 '15 at 1:31










  • $begingroup$
    What may be simpler than $1/n$?
    $endgroup$
    – Michael Galuza
    Jul 14 '15 at 1:31










  • $begingroup$
    You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
    $endgroup$
    – Alex R.
    Jul 14 '15 at 1:33










  • $begingroup$
    Good points. I guess there really is no reason not to use $sum 1/n$.
    $endgroup$
    – TorsionSquid
    Jul 14 '15 at 1:40










  • $begingroup$
    In any event: have you seen this?
    $endgroup$
    – J. M. is not a mathematician
    Jul 14 '15 at 2:51














  • 1




    $begingroup$
    What tests have you done? The integral test is usually done pretty early in this sequence.
    $endgroup$
    – rogerl
    Jul 14 '15 at 1:31










  • $begingroup$
    What may be simpler than $1/n$?
    $endgroup$
    – Michael Galuza
    Jul 14 '15 at 1:31










  • $begingroup$
    You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
    $endgroup$
    – Alex R.
    Jul 14 '15 at 1:33










  • $begingroup$
    Good points. I guess there really is no reason not to use $sum 1/n$.
    $endgroup$
    – TorsionSquid
    Jul 14 '15 at 1:40










  • $begingroup$
    In any event: have you seen this?
    $endgroup$
    – J. M. is not a mathematician
    Jul 14 '15 at 2:51








1




1




$begingroup$
What tests have you done? The integral test is usually done pretty early in this sequence.
$endgroup$
– rogerl
Jul 14 '15 at 1:31




$begingroup$
What tests have you done? The integral test is usually done pretty early in this sequence.
$endgroup$
– rogerl
Jul 14 '15 at 1:31












$begingroup$
What may be simpler than $1/n$?
$endgroup$
– Michael Galuza
Jul 14 '15 at 1:31




$begingroup$
What may be simpler than $1/n$?
$endgroup$
– Michael Galuza
Jul 14 '15 at 1:31












$begingroup$
You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
$endgroup$
– Alex R.
Jul 14 '15 at 1:33




$begingroup$
You don't need the integral test to conclude $sum_n 1/n$ diverges, you can prove it using the comparison test.
$endgroup$
– Alex R.
Jul 14 '15 at 1:33












$begingroup$
Good points. I guess there really is no reason not to use $sum 1/n$.
$endgroup$
– TorsionSquid
Jul 14 '15 at 1:40




$begingroup$
Good points. I guess there really is no reason not to use $sum 1/n$.
$endgroup$
– TorsionSquid
Jul 14 '15 at 1:40












$begingroup$
In any event: have you seen this?
$endgroup$
– J. M. is not a mathematician
Jul 14 '15 at 2:51




$begingroup$
In any event: have you seen this?
$endgroup$
– J. M. is not a mathematician
Jul 14 '15 at 2:51










5 Answers
5






active

oldest

votes


















3












$begingroup$

The simplest one I can think of:



$$ begin{align} S& = 1 + frac12 + frac12 + frac13 + frac13 + frac13 + frac14 + frac14 + frac14 + frac14 + cdots \
&= 1 + ,1quad +quad ,, 1 quad quad, +quad quad ,, 1, + cdots
end{align}
$$



In fact why don't you let your students make the suggestions?






share|cite|improve this answer











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  • $begingroup$
    Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
    $endgroup$
    – Hexacoordinate-C
    Jul 14 '15 at 1:51



















2












$begingroup$

Consider the series
$$
1+frac{1}{2}+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{8}+cdots
$$
(here $frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.



(This is quite similar to the comparison test which proves $sumfrac{1}{n}$ diverges).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
    $endgroup$
    – ra1nmaster
    Jul 14 '15 at 1:39



















2












$begingroup$

The integral test isn't needed: we can argue by contradiction.



Let $S_n = sum_{k=1}^n 1/k$. Then:



$$S_{2n} - S_n= sum_{k=1}^{2n} 1/k - sum_{k=1}^n1/k = sum_{k=n+1}^{2n}1/k$$



If $(S_n)$ converges, then $lim(S_{2n} - S_n) = 0$. However,



$$k le 2n implies frac{1}{2n} le frac{1}{k}$$



Applying sums from $n+1$ till $2n$:



$$frac12 le S_{2n} - S_n$$



Taking $n to infty$, $lim(S_{2n} - S_n) ge 1/2$. Contradiction,






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:



    Suppose $sum^{infty } _{n=1}frac{1}{n}$ converges to a finite number $L$. Then,



    $L=1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+cdots +cdots$ and this is greater than



    $1+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{6}+frac{1}{6}+frac{1}{8}+frac{1}{8}+cdots$. Now group the terms:



    $1+frac{1}{2}+left ( frac{1}{4}+frac{1}{4} right )+left ( frac{1}{6}+frac{1}{6} right )+left ( frac{1}{8}+frac{1}{8} right )+cdots $ and this latter item is



    $1+frac{1}{2}+frac{1}{2}+frac{1}{3}+frac{1}{4}+cdots =L+frac{1}{2}$ so what we've shown is



    $Lgeq L+frac{1}{2}$ which is absurd. Hence, the series does not converge.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      For the harmonic series,
      do like Cauchy's condensation test:



      $begin{array}\
      2 text{ terms } ge 1/4
      &implies sum ge frac12\
      4 text{ terms } ge 1/8
      &implies sum ge frac12\
      8 text{ terms } ge 1/16
      &implies sum ge frac12\
      ...\
      2^n text{ terms } ge 1/2^{n+1}
      &implies sum ge frac12\
      end{array}
      $






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        The simplest one I can think of:



        $$ begin{align} S& = 1 + frac12 + frac12 + frac13 + frac13 + frac13 + frac14 + frac14 + frac14 + frac14 + cdots \
        &= 1 + ,1quad +quad ,, 1 quad quad, +quad quad ,, 1, + cdots
        end{align}
        $$



        In fact why don't you let your students make the suggestions?






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
          $endgroup$
          – Hexacoordinate-C
          Jul 14 '15 at 1:51
















        3












        $begingroup$

        The simplest one I can think of:



        $$ begin{align} S& = 1 + frac12 + frac12 + frac13 + frac13 + frac13 + frac14 + frac14 + frac14 + frac14 + cdots \
        &= 1 + ,1quad +quad ,, 1 quad quad, +quad quad ,, 1, + cdots
        end{align}
        $$



        In fact why don't you let your students make the suggestions?






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
          $endgroup$
          – Hexacoordinate-C
          Jul 14 '15 at 1:51














        3












        3








        3





        $begingroup$

        The simplest one I can think of:



        $$ begin{align} S& = 1 + frac12 + frac12 + frac13 + frac13 + frac13 + frac14 + frac14 + frac14 + frac14 + cdots \
        &= 1 + ,1quad +quad ,, 1 quad quad, +quad quad ,, 1, + cdots
        end{align}
        $$



        In fact why don't you let your students make the suggestions?






        share|cite|improve this answer











        $endgroup$



        The simplest one I can think of:



        $$ begin{align} S& = 1 + frac12 + frac12 + frac13 + frac13 + frac13 + frac14 + frac14 + frac14 + frac14 + cdots \
        &= 1 + ,1quad +quad ,, 1 quad quad, +quad quad ,, 1, + cdots
        end{align}
        $$



        In fact why don't you let your students make the suggestions?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 14 '15 at 1:47

























        answered Jul 14 '15 at 1:40









        corindocorindo

        3,402924




        3,402924












        • $begingroup$
          Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
          $endgroup$
          – Hexacoordinate-C
          Jul 14 '15 at 1:51


















        • $begingroup$
          Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
          $endgroup$
          – Hexacoordinate-C
          Jul 14 '15 at 1:51
















        $begingroup$
        Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
        $endgroup$
        – Hexacoordinate-C
        Jul 14 '15 at 1:51




        $begingroup$
        Your edit make me laugh haha. I think all here understand what you mean lol. Obviously its the best answer possible to do ! ;)
        $endgroup$
        – Hexacoordinate-C
        Jul 14 '15 at 1:51











        2












        $begingroup$

        Consider the series
        $$
        1+frac{1}{2}+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{8}+cdots
        $$
        (here $frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.



        (This is quite similar to the comparison test which proves $sumfrac{1}{n}$ diverges).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
          $endgroup$
          – ra1nmaster
          Jul 14 '15 at 1:39
















        2












        $begingroup$

        Consider the series
        $$
        1+frac{1}{2}+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{8}+cdots
        $$
        (here $frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.



        (This is quite similar to the comparison test which proves $sumfrac{1}{n}$ diverges).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
          $endgroup$
          – ra1nmaster
          Jul 14 '15 at 1:39














        2












        2








        2





        $begingroup$

        Consider the series
        $$
        1+frac{1}{2}+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{8}+cdots
        $$
        (here $frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.



        (This is quite similar to the comparison test which proves $sumfrac{1}{n}$ diverges).






        share|cite|improve this answer









        $endgroup$



        Consider the series
        $$
        1+frac{1}{2}+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{4}+frac{1}{8}+cdots
        $$
        (here $frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.



        (This is quite similar to the comparison test which proves $sumfrac{1}{n}$ diverges).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jul 14 '15 at 1:37









        Michael BurrMichael Burr

        26.9k23262




        26.9k23262












        • $begingroup$
          +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
          $endgroup$
          – ra1nmaster
          Jul 14 '15 at 1:39


















        • $begingroup$
          +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
          $endgroup$
          – ra1nmaster
          Jul 14 '15 at 1:39
















        $begingroup$
        +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
        $endgroup$
        – ra1nmaster
        Jul 14 '15 at 1:39




        $begingroup$
        +1 That's pretty clever, and easy to understand for someone who's never been introduced to the concept.
        $endgroup$
        – ra1nmaster
        Jul 14 '15 at 1:39











        2












        $begingroup$

        The integral test isn't needed: we can argue by contradiction.



        Let $S_n = sum_{k=1}^n 1/k$. Then:



        $$S_{2n} - S_n= sum_{k=1}^{2n} 1/k - sum_{k=1}^n1/k = sum_{k=n+1}^{2n}1/k$$



        If $(S_n)$ converges, then $lim(S_{2n} - S_n) = 0$. However,



        $$k le 2n implies frac{1}{2n} le frac{1}{k}$$



        Applying sums from $n+1$ till $2n$:



        $$frac12 le S_{2n} - S_n$$



        Taking $n to infty$, $lim(S_{2n} - S_n) ge 1/2$. Contradiction,






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          The integral test isn't needed: we can argue by contradiction.



          Let $S_n = sum_{k=1}^n 1/k$. Then:



          $$S_{2n} - S_n= sum_{k=1}^{2n} 1/k - sum_{k=1}^n1/k = sum_{k=n+1}^{2n}1/k$$



          If $(S_n)$ converges, then $lim(S_{2n} - S_n) = 0$. However,



          $$k le 2n implies frac{1}{2n} le frac{1}{k}$$



          Applying sums from $n+1$ till $2n$:



          $$frac12 le S_{2n} - S_n$$



          Taking $n to infty$, $lim(S_{2n} - S_n) ge 1/2$. Contradiction,






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            The integral test isn't needed: we can argue by contradiction.



            Let $S_n = sum_{k=1}^n 1/k$. Then:



            $$S_{2n} - S_n= sum_{k=1}^{2n} 1/k - sum_{k=1}^n1/k = sum_{k=n+1}^{2n}1/k$$



            If $(S_n)$ converges, then $lim(S_{2n} - S_n) = 0$. However,



            $$k le 2n implies frac{1}{2n} le frac{1}{k}$$



            Applying sums from $n+1$ till $2n$:



            $$frac12 le S_{2n} - S_n$$



            Taking $n to infty$, $lim(S_{2n} - S_n) ge 1/2$. Contradiction,






            share|cite|improve this answer









            $endgroup$



            The integral test isn't needed: we can argue by contradiction.



            Let $S_n = sum_{k=1}^n 1/k$. Then:



            $$S_{2n} - S_n= sum_{k=1}^{2n} 1/k - sum_{k=1}^n1/k = sum_{k=n+1}^{2n}1/k$$



            If $(S_n)$ converges, then $lim(S_{2n} - S_n) = 0$. However,



            $$k le 2n implies frac{1}{2n} le frac{1}{k}$$



            Applying sums from $n+1$ till $2n$:



            $$frac12 le S_{2n} - S_n$$



            Taking $n to infty$, $lim(S_{2n} - S_n) ge 1/2$. Contradiction,







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 14 '15 at 1:45







            user230734






























                0












                $begingroup$

                The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:



                Suppose $sum^{infty } _{n=1}frac{1}{n}$ converges to a finite number $L$. Then,



                $L=1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+cdots +cdots$ and this is greater than



                $1+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{6}+frac{1}{6}+frac{1}{8}+frac{1}{8}+cdots$. Now group the terms:



                $1+frac{1}{2}+left ( frac{1}{4}+frac{1}{4} right )+left ( frac{1}{6}+frac{1}{6} right )+left ( frac{1}{8}+frac{1}{8} right )+cdots $ and this latter item is



                $1+frac{1}{2}+frac{1}{2}+frac{1}{3}+frac{1}{4}+cdots =L+frac{1}{2}$ so what we've shown is



                $Lgeq L+frac{1}{2}$ which is absurd. Hence, the series does not converge.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:



                  Suppose $sum^{infty } _{n=1}frac{1}{n}$ converges to a finite number $L$. Then,



                  $L=1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+cdots +cdots$ and this is greater than



                  $1+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{6}+frac{1}{6}+frac{1}{8}+frac{1}{8}+cdots$. Now group the terms:



                  $1+frac{1}{2}+left ( frac{1}{4}+frac{1}{4} right )+left ( frac{1}{6}+frac{1}{6} right )+left ( frac{1}{8}+frac{1}{8} right )+cdots $ and this latter item is



                  $1+frac{1}{2}+frac{1}{2}+frac{1}{3}+frac{1}{4}+cdots =L+frac{1}{2}$ so what we've shown is



                  $Lgeq L+frac{1}{2}$ which is absurd. Hence, the series does not converge.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:



                    Suppose $sum^{infty } _{n=1}frac{1}{n}$ converges to a finite number $L$. Then,



                    $L=1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+cdots +cdots$ and this is greater than



                    $1+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{6}+frac{1}{6}+frac{1}{8}+frac{1}{8}+cdots$. Now group the terms:



                    $1+frac{1}{2}+left ( frac{1}{4}+frac{1}{4} right )+left ( frac{1}{6}+frac{1}{6} right )+left ( frac{1}{8}+frac{1}{8} right )+cdots $ and this latter item is



                    $1+frac{1}{2}+frac{1}{2}+frac{1}{3}+frac{1}{4}+cdots =L+frac{1}{2}$ so what we've shown is



                    $Lgeq L+frac{1}{2}$ which is absurd. Hence, the series does not converge.






                    share|cite|improve this answer









                    $endgroup$



                    The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:



                    Suppose $sum^{infty } _{n=1}frac{1}{n}$ converges to a finite number $L$. Then,



                    $L=1+frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}+cdots +cdots$ and this is greater than



                    $1+frac{1}{2}+frac{1}{4}+frac{1}{4}+frac{1}{6}+frac{1}{6}+frac{1}{8}+frac{1}{8}+cdots$. Now group the terms:



                    $1+frac{1}{2}+left ( frac{1}{4}+frac{1}{4} right )+left ( frac{1}{6}+frac{1}{6} right )+left ( frac{1}{8}+frac{1}{8} right )+cdots $ and this latter item is



                    $1+frac{1}{2}+frac{1}{2}+frac{1}{3}+frac{1}{4}+cdots =L+frac{1}{2}$ so what we've shown is



                    $Lgeq L+frac{1}{2}$ which is absurd. Hence, the series does not converge.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jul 14 '15 at 1:51









                    MatematletaMatematleta

                    11.4k2920




                    11.4k2920























                        0












                        $begingroup$

                        For the harmonic series,
                        do like Cauchy's condensation test:



                        $begin{array}\
                        2 text{ terms } ge 1/4
                        &implies sum ge frac12\
                        4 text{ terms } ge 1/8
                        &implies sum ge frac12\
                        8 text{ terms } ge 1/16
                        &implies sum ge frac12\
                        ...\
                        2^n text{ terms } ge 1/2^{n+1}
                        &implies sum ge frac12\
                        end{array}
                        $






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          For the harmonic series,
                          do like Cauchy's condensation test:



                          $begin{array}\
                          2 text{ terms } ge 1/4
                          &implies sum ge frac12\
                          4 text{ terms } ge 1/8
                          &implies sum ge frac12\
                          8 text{ terms } ge 1/16
                          &implies sum ge frac12\
                          ...\
                          2^n text{ terms } ge 1/2^{n+1}
                          &implies sum ge frac12\
                          end{array}
                          $






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            For the harmonic series,
                            do like Cauchy's condensation test:



                            $begin{array}\
                            2 text{ terms } ge 1/4
                            &implies sum ge frac12\
                            4 text{ terms } ge 1/8
                            &implies sum ge frac12\
                            8 text{ terms } ge 1/16
                            &implies sum ge frac12\
                            ...\
                            2^n text{ terms } ge 1/2^{n+1}
                            &implies sum ge frac12\
                            end{array}
                            $






                            share|cite|improve this answer









                            $endgroup$



                            For the harmonic series,
                            do like Cauchy's condensation test:



                            $begin{array}\
                            2 text{ terms } ge 1/4
                            &implies sum ge frac12\
                            4 text{ terms } ge 1/8
                            &implies sum ge frac12\
                            8 text{ terms } ge 1/16
                            &implies sum ge frac12\
                            ...\
                            2^n text{ terms } ge 1/2^{n+1}
                            &implies sum ge frac12\
                            end{array}
                            $







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jul 14 '15 at 2:25









                            marty cohenmarty cohen

                            74k549128




                            74k549128






























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