Infinite Series Diverges By Divergence Test But Converges By Limit Comparison Test











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I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks



P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$






calculus sequences-and-series convergence divergent-series absolute-convergence







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  • I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
    – Clayton
    Nov 22 at 6:21

















up vote
0
down vote

favorite












Image of My Work



I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks



P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$






calculus sequences-and-series convergence divergent-series absolute-convergence







share|cite|improve this question
























  • I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
    – Clayton
    Nov 22 at 6:21















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Image of My Work



I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks



P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$






calculus sequences-and-series convergence divergent-series absolute-convergence







share|cite|improve this question















Image of My Work



I understand why this infinite series diverges by the divergence test but I can't find fault in my limit comparison test which says it diverges. Please help. Thanks



P.S. if my handwriting threw you off the original equation is
$$sum_{n=0}^{infty} (-1)^n cdot frac{n^4}{n^3 + 1}.$$






calculus sequences-and-series convergence divergent-series absolute-convergence




calculus sequences-and-series convergence divergent-series absolute-convergence






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edited Nov 22 at 6:17









Robert Z

92.1k1058129




92.1k1058129










asked Nov 22 at 6:11









SeanCorc

1




1












  • I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
    – Clayton
    Nov 22 at 6:21




















  • I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
    – Clayton
    Nov 22 at 6:21


















I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21






I think you have a fundamental misunderstanding. First, the limit comparison is only going to work if all of the terms are positive (you might take absolute values, but this changes the problem. In some cases, this change means the difference between convergence and divergence.) Second, the limit of the sequence converges, but the limit of the sum doesn’t converge... At any rate, you compared with $b_n=n$. $sum b_n$ diverges (you might need to think about the statement of the tests to wrap your head around what I mean.)
– Clayton
Nov 22 at 6:21












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There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.






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    Is one of your "diverges" supposed to be "converges"?



    Your sum is essentially
    $sum_{n=1}^{infty} (-1)^nn
    $
    or
    $-1+2-3+4-5+6...$
    with partial sums
    $-1,1,-2,2,-3,3,...$.
    This diverges.



    In what sense can the series converge?



    If you are looking at Cesaro sums,
    the even ones are zero
    and the odd ones
    go to -1/2,
    so these do not converge.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
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      There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.






      share|cite|improve this answer

























        up vote
        1
        down vote













        There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.






          share|cite|improve this answer












          There's nothing wrong with your calculations. You did well to find out the value of $b_n=x$, but it actually diverges, so when you compute the limit of your series over $b_n$, finding it to be 1 actually means that it diverges rather than converges.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 6:23









          Muchang Bahng

          562




          562






















              up vote
              0
              down vote













              Is one of your "diverges" supposed to be "converges"?



              Your sum is essentially
              $sum_{n=1}^{infty} (-1)^nn
              $
              or
              $-1+2-3+4-5+6...$
              with partial sums
              $-1,1,-2,2,-3,3,...$.
              This diverges.



              In what sense can the series converge?



              If you are looking at Cesaro sums,
              the even ones are zero
              and the odd ones
              go to -1/2,
              so these do not converge.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Is one of your "diverges" supposed to be "converges"?



                Your sum is essentially
                $sum_{n=1}^{infty} (-1)^nn
                $
                or
                $-1+2-3+4-5+6...$
                with partial sums
                $-1,1,-2,2,-3,3,...$.
                This diverges.



                In what sense can the series converge?



                If you are looking at Cesaro sums,
                the even ones are zero
                and the odd ones
                go to -1/2,
                so these do not converge.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Is one of your "diverges" supposed to be "converges"?



                  Your sum is essentially
                  $sum_{n=1}^{infty} (-1)^nn
                  $
                  or
                  $-1+2-3+4-5+6...$
                  with partial sums
                  $-1,1,-2,2,-3,3,...$.
                  This diverges.



                  In what sense can the series converge?



                  If you are looking at Cesaro sums,
                  the even ones are zero
                  and the odd ones
                  go to -1/2,
                  so these do not converge.






                  share|cite|improve this answer












                  Is one of your "diverges" supposed to be "converges"?



                  Your sum is essentially
                  $sum_{n=1}^{infty} (-1)^nn
                  $
                  or
                  $-1+2-3+4-5+6...$
                  with partial sums
                  $-1,1,-2,2,-3,3,...$.
                  This diverges.



                  In what sense can the series converge?



                  If you are looking at Cesaro sums,
                  the even ones are zero
                  and the odd ones
                  go to -1/2,
                  so these do not converge.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 6:34









                  marty cohen

                  71.9k547126




                  71.9k547126






























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