Suppose $limlimits_{n rightarrow infty } n^2 a_n =1$, then $ sumlimits _{n=1} ^{infty} a_n$ is convergent? Or...












2












$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31


















2












$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31
















2












2








2





$begingroup$



Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





Suppose $lim_{n rightarrow infty } n^2 a_n =1$ then $ sum _{n=1} ^{infty} a_n$ is convergent or divergent?




We use the definition of the limit, then for $varepsilon =1$ there must exist an $n_0geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that:
$$ 0 < n^2 a_n < 2 $$
We divide by:
$$ 0 < a_n < frac{2}{n^2} $$
We now know if we take the infinite sum we get:
$$ 0 < sum _{n=1} ^{infty} a_n < sum _{n=1} ^{infty} frac{2}{n^2} $$
We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 15:12







Wesley Strik

















asked Dec 1 '18 at 14:47









Wesley StrikWesley Strik

1,653423




1,653423








  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31
















  • 2




    $begingroup$
    you argument is correct
    $endgroup$
    – Guy Fsone
    Dec 1 '18 at 14:52






  • 1




    $begingroup$
    note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
    $endgroup$
    – Masacroso
    Dec 1 '18 at 14:54








  • 1




    $begingroup$
    @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
    $endgroup$
    – Did
    Dec 1 '18 at 15:09










  • $begingroup$
    Of course, it is bounded AND increasing, therefore it converges!
    $endgroup$
    – Wesley Strik
    Dec 1 '18 at 15:11






  • 1




    $begingroup$
    @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
    $endgroup$
    – Did
    Dec 1 '18 at 15:31










2




2




$begingroup$
you argument is correct
$endgroup$
– Guy Fsone
Dec 1 '18 at 14:52




$begingroup$
you argument is correct
$endgroup$
– Guy Fsone
Dec 1 '18 at 14:52




1




1




$begingroup$
note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
$endgroup$
– Masacroso
Dec 1 '18 at 14:54






$begingroup$
note that $a_nge 0$, hence the series converges, that is, the sequence defined by $s_n:=sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem
$endgroup$
– Masacroso
Dec 1 '18 at 14:54






1




1




$begingroup$
@Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
$endgroup$
– Did
Dec 1 '18 at 15:09




$begingroup$
@Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|leqslant c/n^2$ then $sum a_n$ converges.
$endgroup$
– Did
Dec 1 '18 at 15:09












$begingroup$
Of course, it is bounded AND increasing, therefore it converges!
$endgroup$
– Wesley Strik
Dec 1 '18 at 15:11




$begingroup$
Of course, it is bounded AND increasing, therefore it converges!
$endgroup$
– Wesley Strik
Dec 1 '18 at 15:11




1




1




$begingroup$
@Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
$endgroup$
– Did
Dec 1 '18 at 15:31






$begingroup$
@Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $sum b_n$ diverges hence one should not write things like $sumlimits_{n=0}^infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $sum b_n$, which always exists, and the number $sumlimits_{n=0}^infty b_n$, which may or may not exist...)
$endgroup$
– Did
Dec 1 '18 at 15:31












4 Answers
4






active

oldest

votes


















2












$begingroup$

Option.



$lim_{n rightarrow infty}n^2a_n=1.$



This Implies for $n in mathbb{Z^+}$ :



$n^2|a_n| lt M$, positive real number.



$|a_n| lt M/n^2.$



By comparison test $sum |a_n|$ is convergent,



hence $sum a_n$ is convergent.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



      Now two series with (eventually positive) equivalent terms both converge or both diverge.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        It is increasing and bounded, therefore it converges!
        $endgroup$
        – Wesley Strik
        Dec 1 '18 at 15:10






      • 1




        $begingroup$
        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
        $endgroup$
        – Bernard
        Dec 1 '18 at 15:16



















      1












      $begingroup$

      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



      indeed



      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



      therefore $sum a_n$ converges.



      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        The test only applies to positive series, so a small step is needed before applying it.
        $endgroup$
        – Andrés E. Caicedo
        Dec 1 '18 at 15:10






      • 1




        $begingroup$
        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
        $endgroup$
        – gimusi
        Dec 1 '18 at 15:20











      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Option.



      $lim_{n rightarrow infty}n^2a_n=1.$



      This Implies for $n in mathbb{Z^+}$ :



      $n^2|a_n| lt M$, positive real number.



      $|a_n| lt M/n^2.$



      By comparison test $sum |a_n|$ is convergent,



      hence $sum a_n$ is convergent.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Option.



        $lim_{n rightarrow infty}n^2a_n=1.$



        This Implies for $n in mathbb{Z^+}$ :



        $n^2|a_n| lt M$, positive real number.



        $|a_n| lt M/n^2.$



        By comparison test $sum |a_n|$ is convergent,



        hence $sum a_n$ is convergent.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Option.



          $lim_{n rightarrow infty}n^2a_n=1.$



          This Implies for $n in mathbb{Z^+}$ :



          $n^2|a_n| lt M$, positive real number.



          $|a_n| lt M/n^2.$



          By comparison test $sum |a_n|$ is convergent,



          hence $sum a_n$ is convergent.






          share|cite|improve this answer









          $endgroup$



          Option.



          $lim_{n rightarrow infty}n^2a_n=1.$



          This Implies for $n in mathbb{Z^+}$ :



          $n^2|a_n| lt M$, positive real number.



          $|a_n| lt M/n^2.$



          By comparison test $sum |a_n|$ is convergent,



          hence $sum a_n$ is convergent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 15:49









          Peter SzilasPeter Szilas

          10.9k2721




          10.9k2721























              3












              $begingroup$

              A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.






                  share|cite|improve this answer









                  $endgroup$



                  A series converges iff its remainder term $sumlimits_N^infty a_n rightarrow 0$ as $Nrightarrow infty$. Your argument shows that you can, for large enough $N$, bound $|sumlimits_N^infty a_n|$ by $sumlimits_N^infty frac{1}{n^2}$. The latter term approaches zero as $Nrightarrow infty$, since $sum frac{1}{n^2}$ converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 14:56









                  user25959user25959

                  1,573816




                  1,573816























                      2












                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16
















                      2












                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16














                      2












                      2








                      2





                      $begingroup$

                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.






                      share|cite|improve this answer









                      $endgroup$



                      The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $dfrac1{n^2}$.



                      Now two series with (eventually positive) equivalent terms both converge or both diverge.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 '18 at 15:07









                      BernardBernard

                      119k639112




                      119k639112












                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16


















                      • $begingroup$
                        It is increasing and bounded, therefore it converges!
                        $endgroup$
                        – Wesley Strik
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                        $endgroup$
                        – Bernard
                        Dec 1 '18 at 15:16
















                      $begingroup$
                      It is increasing and bounded, therefore it converges!
                      $endgroup$
                      – Wesley Strik
                      Dec 1 '18 at 15:10




                      $begingroup$
                      It is increasing and bounded, therefore it converges!
                      $endgroup$
                      – Wesley Strik
                      Dec 1 '18 at 15:10




                      1




                      1




                      $begingroup$
                      Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                      $endgroup$
                      – Bernard
                      Dec 1 '18 at 15:16




                      $begingroup$
                      Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence.
                      $endgroup$
                      – Bernard
                      Dec 1 '18 at 15:16











                      1












                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20
















                      1












                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20














                      1












                      1








                      1





                      $begingroup$

                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.






                      share|cite|improve this answer











                      $endgroup$



                      Your argument is correct, more simply we can refer directly to limit comparison test with $$sum frac1{n^2}$$



                      indeed



                      $$frac{a_n}{ frac1{n^2}}=n^2a_nto 1$$



                      therefore $sum a_n$ converges.



                      It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.



                      Therefore if you are not requested to use esplicitely the $epsilon-delta$ definition a solution by limit comparison test is fine.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 1 '18 at 14:57

























                      answered Dec 1 '18 at 14:49









                      gimusigimusi

                      1




                      1








                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20














                      • 1




                        $begingroup$
                        The test only applies to positive series, so a small step is needed before applying it.
                        $endgroup$
                        – Andrés E. Caicedo
                        Dec 1 '18 at 15:10






                      • 1




                        $begingroup$
                        @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                        $endgroup$
                        – gimusi
                        Dec 1 '18 at 15:20








                      1




                      1




                      $begingroup$
                      The test only applies to positive series, so a small step is needed before applying it.
                      $endgroup$
                      – Andrés E. Caicedo
                      Dec 1 '18 at 15:10




                      $begingroup$
                      The test only applies to positive series, so a small step is needed before applying it.
                      $endgroup$
                      – Andrés E. Caicedo
                      Dec 1 '18 at 15:10




                      1




                      1




                      $begingroup$
                      @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                      $endgroup$
                      – gimusi
                      Dec 1 '18 at 15:20




                      $begingroup$
                      @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large.
                      $endgroup$
                      – gimusi
                      Dec 1 '18 at 15:20


















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