Is the following argument legal? Say we have $\sum_{n=0}^{\infty}a_{n}$ that absolutely converges and $\sum_{n=0}^{\infty}b_{n}$ converges. Say we also have $f$ that is continuous at $x=0$. I have to prove $\sum_{n=0}^{\infty}a_{n}f(b_{n})$ absolutely converges. Now my idea was - $f$ is continuous at $0$, so we have $\lim_{n\to\infty}f(b_{n})=f(0)$.
Let's denote $S_{p}=\sum_{n=1}^{p}a_{n}f(b_{n})$.
Now, we can say $\lim_{p\to\infty}\sum_{n=1}^{p}a_{n}f(b_{n})=f(0)\lim_{p\to\infty}\sum_{n=1}^{p}a_{n}$.
We know $\lim_{p\to\infty}\sum_{n=1}^{p}a_{n}$ exists because $\sum_{n=1}^{\infty}a_{n}$ converges, so the original limit exists. Is this argument valid?
Or what other way is there to approach such problem?

You must show that $\sum_n |a_n f(b_n)| < \infty$ (this is absolute convergence) and your attempt does not say anything about this. Here is how I would approach this problem:
Since $\sum_n b_n$ converges, we have $b_n \to 0$. Consequently since $f$ is continuous in $0$, $f(b_n) \to f(0)$. In particular, $(f(b_n))_n$ is a bounded sequence so there is $M \geq 0$ such that $|f(b_n)| \leq M$ for all $n \geq 0$. Then
$$\sum_n |a_n f(b_n)| \leq \sum_n |a_n| M = M \sum_n |a_n|< \infty$$
and we can conclude.
