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Given two $a_n,\; b_n>0$ such that $lim_{n \to \infty} a_n,\; b_n=0$ and $\lim_{n \to \infty} \frac{a_n}{b_n}=1$, where neither series is necessarily monotonic: if $\displaystyle \sum_{k=1}^\infty (-1)^k b_k$ converges, does $\displaystyle \sum_{k=1}^\infty (-1)^k a_k$ necessarily converge?

Similarly, given two series $a_n,\; b_n>0$ such that $lim_{n \to \infty} a_n,\; b_n=0$ and $a_n<b_n$ for all $n$ where neither series is necessarily monotonic: if $\displaystyle \sum_{k=1}^\infty (-1)^k b_k$ converges, does $\displaystyle \sum_{k=1}^\infty (-1)^k a_k$ necessarily converge?

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The answer is No in both cases.

Consider $a_k=\dfrac{1}{\sqrt{k}-(-1)^k}$. Clearly $a_k>0$ for $k\geq 1$. Now, $$(-1)^ka_k-\frac{(-1)^k}{\sqrt{k}}=\frac{1}{\sqrt{k}(\sqrt{k}-(-1)^k)}\sim\frac{1}{k} $$ This proves that the series $\displaystyle \sum_{k=1}^\infty\left((-1)^ka_k-\frac{(-1)^k}{\sqrt{k}}\right)$ is divergent, and since $\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}} $ is convergent by the alternating series test we conclude that $\displaystyle \sum_{k=1}^\infty (-1)^ka_k $ is divergent.

Now, let $b_k=\frac{1}{\sqrt{k}-2}$ for $k>4$ (with for example $b_k=3$ for $k=1,2,3,4$). Then it is clear that $\lim\limits_{k\to\infty}\dfrac{a_k}{b_k}=1$ and also it is not hard to show that $a_k<b_k$ for $k\ge 1$. Moreover, by alternating series test, the series $\displaystyle \sum_{k=1}^\infty (-1)^kb_k $ is convergent.

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