Convergence of a cosine series

Question

Given that $$\sum_{n=1}^\infty a_n<\infty,$$ and that $$\lim_{n\to \infty}b_n=0$$ Is the series $$\sum_{n=0}^\infty a_nb_n^{-2}(1-cos(b_n))$$ necessarily convergent?

Answer 1

If the sequence $(a_n)_n$ is a positive sequence, then the answer is yes, since you can apply the Limit–Comparison Test (in a comment I asked whether you assume that the sequence $(a_n)_n$ is positive, since that case is a great exercise about the application of the Limit–Comparison Test; I will gladly detail this part if you answer accordingly).

In the general case, the answer is no, as shown by the following counter-example:

Consider the sequence $(a_n)_{n\in\mathbb{N}^*}$ defined by $$\forall n\in\mathbb{N}^*,\ a_n=\frac{(-1)^n}{n^{1/4}}.$$ It is well-known that the series $\sum_n a_n$ converges (alternating Riemann series).

Now, since the function $$f:[0,2\pi]\longrightarrow[0,1/2]:x\longmapsto\frac{1-\cos(x)}{x^2}$$ is a decreasing bijection (details left to the reader), we can define the sequence $(b_n)_{n\in\mathbb{N}^*}$ as $$\forall n\in\mathbb{N}^*,\ b_n=f^{-1}\left(\frac12-\frac1{a_n}\ln\left(1+\frac{(-1)^n}{4\sqrt n}\right)\right).$$ The sequence $(b_n)_{n\in\mathbb{N}^*}$ is well-defined and its limit is $0$ (details left to the reader). Now, for all $n\in\mathbb{N}^*$, \begin{align*} a_n\frac{1-\cos(b_n)}{b_n^2}&=a_n\,f(b_n)\\ &=\frac{a_n}2-\ln\left(1+\frac{(-1)^n}{4\sqrt n}\right)\\ &\underset{n\to+\infty}=\frac{(-1)^n}{2n^{1/4}}+\frac{(-1)^{n+1}}{4\sqrt n}+\frac1{32n}+o\left(\frac1n\right), \end{align*} and it is easy to check that this is the general term of a divergent series: the two first terms are the general terms of convergent alternating series, and the remaining term, namely, $$\frac1{32n}+o\left(\frac1n\right)\underset{n\to+\infty}\sim\frac1{32n}>0$$ is the general term of a divergent series (keeps a constant sign, hence the Limit–Comparison Test applies).

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The previous counterexample (hopefully) explains how such series can be constructed. A more straightforward counterexample is given by $$a_n=\frac{(-1)^n}{n^{1/4}}$$ and $$b_n=\frac1{n^{1/4}}+\frac{(-1)^n}{\sqrt n}.$$ Then \begin{align*} \frac{1-\cos(b_n)}{b_n^2} &\underset{n\to+\infty}=\frac12-\frac{b_n^2}{24}+o\bigl(b_n^3\bigr)\\ &\underset{n\to+\infty}=\frac12-\frac1{24\sqrt n}-\frac{(-1)^n}{12n^{3/4}}+o\left(\frac1{n^{3/4}}\right) \end{align*} hence \begin{align*} a_n\frac{1-\cos(b_n)}{b_n^2} &\underset{n\to+\infty}=\frac{(-1)^n}{2n^{1/4}}-\frac{(-1)^n}{24n^{3/4}}-\frac1{12n}+o\left(\frac1n\right), \end{align*} which is the general term of a divergent series.