Absolute or conditional convergence?












0














Determine whether the series:



$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$



converges absolutely, conditionally or diverges.



I know the series converges conditionally using alternating series test.



My question is how do I determine absolute convergence here.



I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?






calculus sequences-and-series absolute-convergence conditional-convergence







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  • 1




    How did your comparison test result in only conditional convergence?
    – Arthur
    Nov 26 at 23:14


















0














Determine whether the series:



$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$



converges absolutely, conditionally or diverges.



I know the series converges conditionally using alternating series test.



My question is how do I determine absolute convergence here.



I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?






calculus sequences-and-series absolute-convergence conditional-convergence







share|cite|improve this question


















  • 1




    How did your comparison test result in only conditional convergence?
    – Arthur
    Nov 26 at 23:14
















0












0








0







Determine whether the series:



$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$



converges absolutely, conditionally or diverges.



I know the series converges conditionally using alternating series test.



My question is how do I determine absolute convergence here.



I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?






calculus sequences-and-series absolute-convergence conditional-convergence







share|cite|improve this question













Determine whether the series:



$$sum_{n=1}^infty (-1)^n frac {2n^2+3n+4} {2n^4 + 3}$$



converges absolutely, conditionally or diverges.



I know the series converges conditionally using alternating series test.



My question is how do I determine absolute convergence here.



I tried limit comparison test with $frac{1}{n^2}$ which results in conditional convergence only, is that the way to go or not?






calculus sequences-and-series absolute-convergence conditional-convergence




calculus sequences-and-series absolute-convergence conditional-convergence






share|cite|improve this question













share|cite|improve this question











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asked Nov 26 at 23:11









J. Lastin

1025




1025








  • 1




    How did your comparison test result in only conditional convergence?
    – Arthur
    Nov 26 at 23:14
















  • 1




    How did your comparison test result in only conditional convergence?
    – Arthur
    Nov 26 at 23:14










1




1




How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14






How did your comparison test result in only conditional convergence?
– Arthur
Nov 26 at 23:14












1 Answer
1






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oldest

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3














Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.






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    Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.






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      Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.






      share|cite|improve this answer
























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        Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.






        share|cite|improve this answer












        Since$$leftlvert(-1)^nfrac{2n^2+3n+4}{2n^4+3}rightrvert=frac{2n^2+3n+4}{2n^4+3}$$and since$$lim_{ntoinfty}frac{frac{2n^2+3n+4}{2n^4+3}}{frac1{n^2}}=1,$$one deduces from the comparaison test that the series converges absolutely.







        share|cite|improve this answer












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        answered Nov 26 at 23:17









        José Carlos Santos

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