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












-3














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







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put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago


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




    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 $n to infty$.
    – Olivier Oloa
    2 days ago












  • @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    – Theodossis Papadopoulos
    2 days ago












  • Do you know when $sum frac1{n^a}$ converges?
    – Olivier Oloa
    2 days ago






  • 1




    Thus your initial series is divergent, by the limit comparison test.
    – Olivier Oloa
    2 days ago








  • 1




    @OlivierOloa thanks for clearing things out!
    – Theodossis Papadopoulos
    2 days ago
















-3














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







New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago


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




    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 $n to infty$.
    – Olivier Oloa
    2 days ago












  • @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    – Theodossis Papadopoulos
    2 days ago












  • Do you know when $sum frac1{n^a}$ converges?
    – Olivier Oloa
    2 days ago






  • 1




    Thus your initial series is divergent, by the limit comparison test.
    – Olivier Oloa
    2 days ago








  • 1




    @OlivierOloa thanks for clearing things out!
    – Theodossis Papadopoulos
    2 days ago














-3












-3








-3


1





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







New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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







New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Theodossis PapadopoulosTheodossis Papadopoulos

186




186




New contributor




Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago


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.




put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago


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




    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 $n to infty$.
    – Olivier Oloa
    2 days ago












  • @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    – Theodossis Papadopoulos
    2 days ago












  • Do you know when $sum frac1{n^a}$ converges?
    – Olivier Oloa
    2 days ago






  • 1




    Thus your initial series is divergent, by the limit comparison test.
    – Olivier Oloa
    2 days ago








  • 1




    @OlivierOloa thanks for clearing things out!
    – Theodossis Papadopoulos
    2 days ago














  • 2




    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 $n to infty$.
    – Olivier Oloa
    2 days ago












  • @OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
    – Theodossis Papadopoulos
    2 days ago












  • Do you know when $sum frac1{n^a}$ converges?
    – Olivier Oloa
    2 days ago






  • 1




    Thus your initial series is divergent, by the limit comparison test.
    – Olivier Oloa
    2 days ago








  • 1




    @OlivierOloa thanks for clearing things out!
    – Theodossis Papadopoulos
    2 days ago








2




2




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 $n to infty$.
– Olivier Oloa
2 days ago






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 $n to infty$.
– Olivier Oloa
2 days ago














@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago






@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago














Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago




Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago




1




1




Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago






Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago






1




1




@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago




@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago










3 Answers
3






active

oldest

votes


















1














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





























    2














    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





























      1














      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




























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        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


























          1














          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
























            1












            1








            1






            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












            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 2 days ago









            Olivier OloaOlivier Oloa

            108k17176293




            108k17176293























                2














                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


























                  2














                  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
























                    2












                    2








                    2






                    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












                    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 2 days ago









                    Mostafa AyazMostafa Ayaz

                    14.1k3937




                    14.1k3937























                        1














                        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


























                          1














                          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
























                            1












                            1








                            1






                            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












                            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 2 days ago









                            BernardBernard

                            118k639112




                            118k639112















                                -

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