Convergence of a cosine series 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?
 A: 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).

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.
