Check if series converges $sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$ [closed]












-3












$begingroup$


Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$

A quick hint will help me out!






sequences-and-series convergence







share|cite|improve this question









$endgroup$



closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, max_zorn, metamorphy, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:40












  • $begingroup$
    @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:42












  • $begingroup$
    Do you know when $sum frac1{n^a}$ converges?
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:42






  • 1




    $begingroup$
    Thus your initial series is divergent, by the limit comparison test.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:47








  • 1




    $begingroup$
    @OlivierOloa thanks for clearing things out!
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:48
















-3












$begingroup$


Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$

A quick hint will help me out!






sequences-and-series convergence







share|cite|improve this question









$endgroup$



closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, max_zorn, metamorphy, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:40












  • $begingroup$
    @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:42












  • $begingroup$
    Do you know when $sum frac1{n^a}$ converges?
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:42






  • 1




    $begingroup$
    Thus your initial series is divergent, by the limit comparison test.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:47








  • 1




    $begingroup$
    @OlivierOloa thanks for clearing things out!
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:48














-3












-3








-3


1



$begingroup$


Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$

A quick hint will help me out!






sequences-and-series convergence







share|cite|improve this question









$endgroup$




Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$

A quick hint will help me out!






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 13:34









Theodossis PapadopoulosTheodossis Papadopoulos

186




186




closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, max_zorn, metamorphy, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Saad, max_zorn, metamorphy, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:40












  • $begingroup$
    @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:42












  • $begingroup$
    Do you know when $sum frac1{n^a}$ converges?
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:42






  • 1




    $begingroup$
    Thus your initial series is divergent, by the limit comparison test.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:47








  • 1




    $begingroup$
    @OlivierOloa thanks for clearing things out!
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:48














  • 2




    $begingroup$
    Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:40












  • $begingroup$
    @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:42












  • $begingroup$
    Do you know when $sum frac1{n^a}$ converges?
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:42






  • 1




    $begingroup$
    Thus your initial series is divergent, by the limit comparison test.
    $endgroup$
    – Olivier Oloa
    Jan 5 at 13:47








  • 1




    $begingroup$
    @OlivierOloa thanks for clearing things out!
    $endgroup$
    – Theodossis Papadopoulos
    Jan 5 at 13:48








2




2




$begingroup$
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40






$begingroup$
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40














$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42






$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42














$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42




$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42




1




1




$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47






$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47






1




1




$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48




$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48










3 Answers
3






active

oldest

votes


















1












$begingroup$

As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint



    For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Without fiddling with factorisation, &c., it' simpler to use equivalents:



      A polynomial is asymptotically equivalent to its leading term, so
      $$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
      and the latter is the general term of a divergent $p$-series.






      share|cite|improve this answer









      $endgroup$




















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        As $n to infty$, one has
        $$
        frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
        $$         then use the limit comparison test.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          As $n to infty$, one has
          $$
          frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
          $$           then use the limit comparison test.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            As $n to infty$, one has
            $$
            frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
            $$             then use the limit comparison test.






            share|cite|improve this answer









            $endgroup$



            As $n to infty$, one has
            $$
            frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
            $$             then use the limit comparison test.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 13:49









            Olivier OloaOlivier Oloa

            108k17176293




            108k17176293























                2












                $begingroup$

                Hint



                For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Hint



                  For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Hint



                    For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$






                    share|cite|improve this answer









                    $endgroup$



                    Hint



                    For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 5 at 14:43









                    Mostafa AyazMostafa Ayaz

                    14.8k3938




                    14.8k3938























                        1












                        $begingroup$

                        Without fiddling with factorisation, &c., it' simpler to use equivalents:



                        A polynomial is asymptotically equivalent to its leading term, so
                        $$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
                        and the latter is the general term of a divergent $p$-series.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Without fiddling with factorisation, &c., it' simpler to use equivalents:



                          A polynomial is asymptotically equivalent to its leading term, so
                          $$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
                          and the latter is the general term of a divergent $p$-series.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Without fiddling with factorisation, &c., it' simpler to use equivalents:



                            A polynomial is asymptotically equivalent to its leading term, so
                            $$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
                            and the latter is the general term of a divergent $p$-series.






                            share|cite|improve this answer









                            $endgroup$



                            Without fiddling with factorisation, &c., it' simpler to use equivalents:



                            A polynomial is asymptotically equivalent to its leading term, so
                            $$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
                            and the latter is the general term of a divergent $p$-series.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 14:03









                            BernardBernard

                            119k639112




                            119k639112















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