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The conditions for this test state that the terms have to be greater than zero and decreasing, but looking at proofs of this test it isn't clear to me why the terms cannot equal zero.

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  • $\begingroup$ Of course the terms go to $0$ in the limit, that's a necessary (but insufficient) condition for the convergence of any series. Otherwise, you need the terms to decrease (in absolute value) to $0$ in order to invoke Leibniz. The test fails generally, though there may well be other series which you can approach in a similar manner. $\endgroup$
    – lulu
    Apr 11, 2022 at 13:20
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    $\begingroup$ What is a “term at infinity”? $\endgroup$
    – Martin R
    Apr 11, 2022 at 13:21
  • $\begingroup$ You cannot evaluate a term in a series at infinity, you can just look at its limit. The idea is that it will alternate and approach zero from both positive and negative $\endgroup$
    – Henry Lee
    Apr 11, 2022 at 13:21
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    $\begingroup$ If one of the terms is equal to zero, then, because we assume the sequence to be decreasing, the rest will be equal to zero as well. Hence the series will be a finite sum, and any convergence result becomes uninteresting. $\endgroup$ Apr 11, 2022 at 13:21
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    $\begingroup$ One issue here is that your question is very vague. Try to precisely write out a test, generalizing Leibniz, of the form you want. Given that it should be easy either to prove that your test is valid or to exhibit a counterexample. $\endgroup$
    – lulu
    Apr 11, 2022 at 13:28

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Yes, you can have $a_N=0$ for some $N\in\Bbb N$. But then, since the sequence $(a_n)_{n\in\Bbb N}$ is decreasing, you have $a_n=0$ when $n\geqslant N$. And in that case the test is quite uninteresting.

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