If $sum_{n=1}^{infty} a_n$ converges, does $sum_{n=1}^{infty}frac{1}{a_n}$ diverge?












-2












$begingroup$


If a series $sum_{n=1}^{infty} a_n$ converges, does $sum_{n=1}^{infty} dfrac{1}{a_n}$ diverge to infinity?






sequences-and-series convergence divergent-series







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$endgroup$








  • 4




    $begingroup$
    It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
    $endgroup$
    – lulu
    Dec 27 '18 at 15:32










  • $begingroup$
    Does $sum_{n=1}^infty (-1)^n n$ converge?
    $endgroup$
    – Andrei
    Dec 27 '18 at 15:35






  • 1




    $begingroup$
    @Andrei Does it diverge to $infty$?
    $endgroup$
    – N. S.
    Dec 27 '18 at 15:48
















-2












$begingroup$


If a series $sum_{n=1}^{infty} a_n$ converges, does $sum_{n=1}^{infty} dfrac{1}{a_n}$ diverge to infinity?






sequences-and-series convergence divergent-series







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
    $endgroup$
    – lulu
    Dec 27 '18 at 15:32










  • $begingroup$
    Does $sum_{n=1}^infty (-1)^n n$ converge?
    $endgroup$
    – Andrei
    Dec 27 '18 at 15:35






  • 1




    $begingroup$
    @Andrei Does it diverge to $infty$?
    $endgroup$
    – N. S.
    Dec 27 '18 at 15:48














-2












-2








-2





$begingroup$


If a series $sum_{n=1}^{infty} a_n$ converges, does $sum_{n=1}^{infty} dfrac{1}{a_n}$ diverge to infinity?






sequences-and-series convergence divergent-series







share|cite|improve this question











$endgroup$




If a series $sum_{n=1}^{infty} a_n$ converges, does $sum_{n=1}^{infty} dfrac{1}{a_n}$ diverge to infinity?






sequences-and-series convergence divergent-series




sequences-and-series convergence divergent-series






share|cite|improve this question















share|cite|improve this question













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edited Dec 27 '18 at 15:39









Blue

48.5k870154




48.5k870154










asked Dec 27 '18 at 15:30









mrMoonpenguinmrMoonpenguin

153




153








  • 4




    $begingroup$
    It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
    $endgroup$
    – lulu
    Dec 27 '18 at 15:32










  • $begingroup$
    Does $sum_{n=1}^infty (-1)^n n$ converge?
    $endgroup$
    – Andrei
    Dec 27 '18 at 15:35






  • 1




    $begingroup$
    @Andrei Does it diverge to $infty$?
    $endgroup$
    – N. S.
    Dec 27 '18 at 15:48














  • 4




    $begingroup$
    It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
    $endgroup$
    – lulu
    Dec 27 '18 at 15:32










  • $begingroup$
    Does $sum_{n=1}^infty (-1)^n n$ converge?
    $endgroup$
    – Andrei
    Dec 27 '18 at 15:35






  • 1




    $begingroup$
    @Andrei Does it diverge to $infty$?
    $endgroup$
    – N. S.
    Dec 27 '18 at 15:48








4




4




$begingroup$
It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
$endgroup$
– lulu
Dec 27 '18 at 15:32




$begingroup$
It can oscillate, take $a_n=frac {(-1)^n}n$ for example.
$endgroup$
– lulu
Dec 27 '18 at 15:32












$begingroup$
Does $sum_{n=1}^infty (-1)^n n$ converge?
$endgroup$
– Andrei
Dec 27 '18 at 15:35




$begingroup$
Does $sum_{n=1}^infty (-1)^n n$ converge?
$endgroup$
– Andrei
Dec 27 '18 at 15:35




1




1




$begingroup$
@Andrei Does it diverge to $infty$?
$endgroup$
– N. S.
Dec 27 '18 at 15:48




$begingroup$
@Andrei Does it diverge to $infty$?
$endgroup$
– N. S.
Dec 27 '18 at 15:48










3 Answers
3






active

oldest

votes


















1












$begingroup$

It depends on your definition of "diverge". Some take it as "have a $pminfty$ limit" other take it as "non convergent"



I'll stick with the wikipedia definition (emphasys mine)




In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.



If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.




According to this definition, $sum_{n}frac{1}{a_n}$ clearly diverge if $sum_{n}a_n$ converge. Because $a_n$ and $frac{1}{a_n}$ can't both have $0$ as a limit.



But that doesn't mean that $sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:



$$a_n = frac{{(-1)}^n}{n}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A series converges if and only if the $lim_{n to infty} a(n) = 0$. Therefore, $lim_{n to infty} {1over a(n)} = infty$. Clearly, $lvert a(n)rvert < lvert a(n+1)rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+infty$, $-infty$, or oscillates between both.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      No. In fact, given that $sum a_n$ converges, $sumfrac{1}{a_n}toinfty$ if and only if there is some $N in mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n neq 0$ for all $n$.



      Proof:



      In the forwards direction, since $sum frac{1}{a_n}toinfty$, there is, in particular, some $N inmathbb{N}$ such that for all $n > N$, we have $frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n neq 0$ is obvious. For any $C > 0$, it is also true that $frac{1}{C} > 0$, and $(a_n)$ is null (since $sum a_n$ converges), so there is some $Minmathbb{N}$ such that for all $n > M$, $|a_n| < frac{1}{C}$. Thus, taking $L = max(N,M)$, for any $n > L$, we have $0 < a_n < frac{1}{C}$, so $frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $sumfrac{1}{a_n} > C$, hence the result.






      share|cite|improve this answer









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        Your Answer





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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        It depends on your definition of "diverge". Some take it as "have a $pminfty$ limit" other take it as "non convergent"



        I'll stick with the wikipedia definition (emphasys mine)




        In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.



        If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.




        According to this definition, $sum_{n}frac{1}{a_n}$ clearly diverge if $sum_{n}a_n$ converge. Because $a_n$ and $frac{1}{a_n}$ can't both have $0$ as a limit.



        But that doesn't mean that $sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:



        $$a_n = frac{{(-1)}^n}{n}$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          It depends on your definition of "diverge". Some take it as "have a $pminfty$ limit" other take it as "non convergent"



          I'll stick with the wikipedia definition (emphasys mine)




          In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.



          If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.




          According to this definition, $sum_{n}frac{1}{a_n}$ clearly diverge if $sum_{n}a_n$ converge. Because $a_n$ and $frac{1}{a_n}$ can't both have $0$ as a limit.



          But that doesn't mean that $sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:



          $$a_n = frac{{(-1)}^n}{n}$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            It depends on your definition of "diverge". Some take it as "have a $pminfty$ limit" other take it as "non convergent"



            I'll stick with the wikipedia definition (emphasys mine)




            In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.



            If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.




            According to this definition, $sum_{n}frac{1}{a_n}$ clearly diverge if $sum_{n}a_n$ converge. Because $a_n$ and $frac{1}{a_n}$ can't both have $0$ as a limit.



            But that doesn't mean that $sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:



            $$a_n = frac{{(-1)}^n}{n}$$






            share|cite|improve this answer









            $endgroup$



            It depends on your definition of "diverge". Some take it as "have a $pminfty$ limit" other take it as "non convergent"



            I'll stick with the wikipedia definition (emphasys mine)




            In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.



            If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges.




            According to this definition, $sum_{n}frac{1}{a_n}$ clearly diverge if $sum_{n}a_n$ converge. Because $a_n$ and $frac{1}{a_n}$ can't both have $0$ as a limit.



            But that doesn't mean that $sum_{n}a_n$ as an infinite limit. The series can have no limit at all, as per @lulu example in the comments:



            $$a_n = frac{{(-1)}^n}{n}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 15:44









            F.CaretteF.Carette

            1,21612




            1,21612























                0












                $begingroup$

                A series converges if and only if the $lim_{n to infty} a(n) = 0$. Therefore, $lim_{n to infty} {1over a(n)} = infty$. Clearly, $lvert a(n)rvert < lvert a(n+1)rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+infty$, $-infty$, or oscillates between both.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  A series converges if and only if the $lim_{n to infty} a(n) = 0$. Therefore, $lim_{n to infty} {1over a(n)} = infty$. Clearly, $lvert a(n)rvert < lvert a(n+1)rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+infty$, $-infty$, or oscillates between both.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    A series converges if and only if the $lim_{n to infty} a(n) = 0$. Therefore, $lim_{n to infty} {1over a(n)} = infty$. Clearly, $lvert a(n)rvert < lvert a(n+1)rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+infty$, $-infty$, or oscillates between both.






                    share|cite|improve this answer









                    $endgroup$



                    A series converges if and only if the $lim_{n to infty} a(n) = 0$. Therefore, $lim_{n to infty} {1over a(n)} = infty$. Clearly, $lvert a(n)rvert < lvert a(n+1)rvert$. Because each term is larger (in the absolute sense) than the previous, it diverges to either $+infty$, $-infty$, or oscillates between both.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 27 '18 at 15:48









                    William GrannisWilliam Grannis

                    991521




                    991521























                        0












                        $begingroup$

                        No. In fact, given that $sum a_n$ converges, $sumfrac{1}{a_n}toinfty$ if and only if there is some $N in mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n neq 0$ for all $n$.



                        Proof:



                        In the forwards direction, since $sum frac{1}{a_n}toinfty$, there is, in particular, some $N inmathbb{N}$ such that for all $n > N$, we have $frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n neq 0$ is obvious. For any $C > 0$, it is also true that $frac{1}{C} > 0$, and $(a_n)$ is null (since $sum a_n$ converges), so there is some $Minmathbb{N}$ such that for all $n > M$, $|a_n| < frac{1}{C}$. Thus, taking $L = max(N,M)$, for any $n > L$, we have $0 < a_n < frac{1}{C}$, so $frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $sumfrac{1}{a_n} > C$, hence the result.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          No. In fact, given that $sum a_n$ converges, $sumfrac{1}{a_n}toinfty$ if and only if there is some $N in mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n neq 0$ for all $n$.



                          Proof:



                          In the forwards direction, since $sum frac{1}{a_n}toinfty$, there is, in particular, some $N inmathbb{N}$ such that for all $n > N$, we have $frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n neq 0$ is obvious. For any $C > 0$, it is also true that $frac{1}{C} > 0$, and $(a_n)$ is null (since $sum a_n$ converges), so there is some $Minmathbb{N}$ such that for all $n > M$, $|a_n| < frac{1}{C}$. Thus, taking $L = max(N,M)$, for any $n > L$, we have $0 < a_n < frac{1}{C}$, so $frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $sumfrac{1}{a_n} > C$, hence the result.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            No. In fact, given that $sum a_n$ converges, $sumfrac{1}{a_n}toinfty$ if and only if there is some $N in mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n neq 0$ for all $n$.



                            Proof:



                            In the forwards direction, since $sum frac{1}{a_n}toinfty$, there is, in particular, some $N inmathbb{N}$ such that for all $n > N$, we have $frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n neq 0$ is obvious. For any $C > 0$, it is also true that $frac{1}{C} > 0$, and $(a_n)$ is null (since $sum a_n$ converges), so there is some $Minmathbb{N}$ such that for all $n > M$, $|a_n| < frac{1}{C}$. Thus, taking $L = max(N,M)$, for any $n > L$, we have $0 < a_n < frac{1}{C}$, so $frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $sumfrac{1}{a_n} > C$, hence the result.






                            share|cite|improve this answer









                            $endgroup$



                            No. In fact, given that $sum a_n$ converges, $sumfrac{1}{a_n}toinfty$ if and only if there is some $N in mathbb{N}$ such that for all $n > N$, we have $a_n > 0$ and $a_n neq 0$ for all $n$.



                            Proof:



                            In the forwards direction, since $sum frac{1}{a_n}toinfty$, there is, in particular, some $N inmathbb{N}$ such that for all $n > N$, we have $frac{1}{a_n} > 0$, hence $a_n > 0$. In the reverse direction, the necessity of $a_n neq 0$ is obvious. For any $C > 0$, it is also true that $frac{1}{C} > 0$, and $(a_n)$ is null (since $sum a_n$ converges), so there is some $Minmathbb{N}$ such that for all $n > M$, $|a_n| < frac{1}{C}$. Thus, taking $L = max(N,M)$, for any $n > L$, we have $0 < a_n < frac{1}{C}$, so $frac{1}{a_n} > C$, hence (up to a constant that you can feel free to get rid of for yourself), $sumfrac{1}{a_n} > C$, hence the result.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 27 '18 at 15:49









                            user3482749user3482749

                            4,266919




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