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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 $$\sum_{n=0}^{\infty} (-1)^n \cdot \frac{n^4}{n^3 + 1}.$$

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  • $\begingroup$ 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.) $\endgroup$ – Clayton Nov 22 '18 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.

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