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$\sum\limits_{n=1}^\infty a_n$ is a series such that $a_n>0$ when $n$ is odd and $a_n<0$ when $n$ is even, also the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $0$. If $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|= 1$, does it mean that the series $\sum\limits_{n=1}^\infty a_n$ is convergent?

I am confused, since the ratio test tells us nothing, and we lack of the info about if $a_n\geq a_{n+1}$ to use the alternating series test. Or the limit implies the relation between $a_{n+1}$ and $a_n$? So that we can use the alternating series test to prove the convergence of $\sum\limits_{n=1}^\infty a_n$?

If not, are there any counterexamples?

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    $\begingroup$ @Riemann'sPointyNose we know that an is going to 0 and it is alternating, but we do not know if an is greater than an+1 from what we have. $\endgroup$
    – Lei Yu
    Commented Jul 28, 2020 at 23:52
  • $\begingroup$ Hint: let $b_n$ be a positive, decreasing, sequence so that $\sum b_n$ diverges. let $c_n$ be positive, decaying to zero, but has $\lim \frac{b_n}{c_n} = 0$. Now send $a_{2n+1} = b_n + c_n$ and $a_{2n} = - c_n$. I'll leave it to you to come up with the sequences and prove that they satisfy your conditions. $\endgroup$ Commented Jul 29, 2020 at 1:43

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There is a counterexample.

The sequence $\left\{a_n\right\}_{n\in\mathbb{N}}$ such that

$a_{2m-1}=\frac{1+\ln(m+1)}{(m+1)\ln(m+1)}\;\;$ for all $\;\;m\in\mathbb{N}$,

$a_{2m}=-\frac{1}{m+1}\;\;$ for all $\;\;m\in\mathbb{N}$,

satisfies the following properties:

  1. $\lim_\limits{n\to\infty}a_n=0$,
  2. $\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1$,

but $\sum_\limits{n=1}^\infty a_n$ is not convergent, indeed $s_{2p}=\sum_\limits{n=1}^{2p}a_n=\sum_\limits{m=1}^p \left(a_{2m-1}+a_{2m}\right)=\sum_\limits{m=1}^p \left(\frac{1+\ln(m+1)}{(m+1)\ln(m+1)}-\frac{1}{m+1}\right)=\sum_\limits{m=1}^p\frac{1}{(m+1)\ln(m+1)}\to+\infty\\\text{as }\;p\to\infty.$

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  • $\begingroup$ That surely answers the question. Also, the series $\ln 2\,=\,1\,-\,\frac{1}{2}\,+\,\frac{1}{3}\,-+\,\dots\,\,\,$ shows that under the same conditions, a series may converge. $\endgroup$
    – 311411
    Commented Jul 29, 2020 at 20:15
  • $\begingroup$ Of course, indeed with that hypothesis it is impossible to prove that the series is convergent and it is impossible to prove that the series is not convergent because in any case there are counterexamples. You wrote a counterexample of a possible property that claims that the series in not convergent. If we wish to prove the convergence, we need to add other hypothesis. $\endgroup$
    – Angelo
    Commented Jul 29, 2020 at 20:25

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