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Question:

Assume that $a_{n}\in\mathbb R$, and let the series $$\sum_{n=1}^{\infty}a_{n}$$ be convergent

I would like to prove or disprove the convergence of $$\sum_{n=1}^{\infty}\dfrac{(-1)^n\cdot a_{n}}{n}.$$

If $$a_{n}=\dfrac{1}{n^p},p>1,$$

It is clear that the series $\sum_{n=1}^{\infty}\dfrac{(-1)^n\cdot a_{n}}{n}$ is convergent.

Pierre-Yves Gaillard
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math110
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2 Answers2

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This is false. Take $a_n=\frac{(-1)^n}{\log(n)}$. The first series converge by alternating series test, but the second diverges by Cauchy condensation test. See here for what I mean for Cauchy condensation: Convergence of series involving in iterated logarithms $\sum \frac{1}{n(\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k} }$

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Lost1
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If $a_n$ is a positive-valued succession, then: $$\left|\frac{(-1)^na_n}{n}\right|=\frac{|a_n|}{n}=\frac{a_n}{n}\leq a_n,$$ thus the modified series converges by the comparison test. However, in general, this may not hold. If $a_n$ converges absolutely, the argument still holds. Otherwise, it depends on the form of $a_n$, as shown by the example in the other answer, where $a_n=\frac{(-1)^n}{\log(n)}$ converges by the Leibniz test, whereas $\frac{(-1)^na_n}{n}=\frac{1}{\log(n)}$ diverges, for which cfr this question, all those convergences meant as series, not as successions.

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MickG
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  • @boywholived I'm not sure what you mean. You can always edit things. In any case, my answer first deals with the case $a_n\geq0$, then giving the displayed equation, and then does what it can about the general case, so I don't really see what would need to be corrected. It is clear that, in the question, the statement "If $a_n=$" etc holds if $|a_n|=\frac{1}{n^p}$. – MickG Aug 16 '14 at 10:10