Can the product of the terms of a convergent series and a divergent series be convergent? I am investigating positive series, and have so far proved that the series $S=\Sigma^\infty_{n=1}u_n$ is convergent for both $u_n=\frac{n^{10}}{2^n}$ and $u_n=\frac{n^{10}}{n!}$.
I am now faced with the series where $u_n=\frac{n!}{10^n}$. I noticed that the product of the terms of this series, and of the series where $u_n=\frac{n^{10}}{n!}$, which is convergent, is $\frac{n^{10}}{10^n}$, which must be convergent as the series where $u_n=\frac{n^{10}}{2^n}$ is convergent.
So I was wondering if, if the product of the terms of two series is convergent, and you know that one of the series is convergent, does this necessarily imply that the other series must be convergent? i.e. does my method of proof hold?

It does not hold because $\sum_{n=0}^\infty\frac{n!}{10^n}=+\infty$. To see it notice that for $n\geq900$
$$
n!\geq \prod_{j=10}^{n}j\geq10^9\left(\prod_{j=10,j\not\mid100}^nj\right)\left(\prod_{j\mid100}\frac{j}{10}\right)\geq10^{n}.
$$
I.e. the general term of your series doesn't converge to $0$ so the series must diverge.
