Assume that the products $\prod_{n=1}^{\infty} a_n$ and $\prod_{n=1}^{\infty} b_n$ with positive factors both converge. Study the convergence of convergence of:
(a) $\prod_{n=1}^{\infty} a_n^2$;
(b) $\prod_{n=1}^{\infty} a_nb_n$.
My textbook solves these by looking at $\sum_{n=1}^\infty \log(a_n^2)$, $\sum_{n=1}^\infty \log(a_nb_n)$ and $\sum_{n=1}^\infty \log \frac{a_n}{b_n}$, I tried a different strategy and I would like to ask to you for confirmation or helping me find mistakes.
My solution for (a): since $$\prod_{n=1}^{k} a_n^2=(a_1\cdot a_1)\cdot(a_2 \cdot a_2)\cdot...\cdot(a_k\cdot a_k)=(a_1\cdot a_2\cdot...\cdot a_k)\cdot(a_1\cdot a_2 \cdot ... \cdot a_k)=\left(\prod_{n=1}^{k} a_n\right)\left(\prod_{n=1}^{k} a_n\right)=\left(\prod_{n=1}^{k} a_n\right)^2$$ Taking the limit as $k \to \infty$ both sides and using the continuity of the square power it is $$\prod_{n=1}^{\infty} a_n^2=\lim_{k\to\infty}\prod_{n=1}^{k} a_n^2=\lim_{k\to\infty} \left(\prod_{n=1}^{k} a_n\right)^2=\left(\lim_{k \to\infty}\prod_{n=1}^{k} a_n\right)^2$$ But by hypothesis $\prod_{n=1}^{\infty} a_n$ converges, hence $\prod_{n=1}^{\infty} a_n=l\in[0,\infty)$ and so $$\left(\lim_{k \to\infty}\prod_{n=1}^{k} a_n\right)^2=l^2$$ Hence the product converges to $l^2$.
My solution for (b): since $$\prod_{n=1}^{k} a_n b_n=(a_1\cdot b_1)\cdot(a_2 \cdot b_2)\cdot...\cdot(a_k\cdot b_k)=(a_1\cdot a_2\cdot...\cdot a_k)\cdot(b_1\cdot b_2 \cdot ... \cdot b_k)=\left(\prod_{n=1}^{k} a_n\right)\left(\prod_{n=1}^{k} b_n\right)$$ Taking the limit as $k \to \infty$ both sides and using the theorem "limit of product is product of limits when they converges", it is $$\prod_{n=1}^{\infty} a_n b_n=\lim_{k\to\infty}\prod_{n=1}^{k} a_n b_n=\lim_{k\to\infty} \left[\left(\prod_{n=1}^{k} a_n\right)\left(\prod_{n=1}^{k} b_n\right)\right]=\left(\lim_{k\to\infty}\prod_{n=1}^{k} a_n\right)\cdot\left(\lim_{k \to \infty}\prod_{n=1}^{k} b_n\right)$$ But by hypothesis $\prod_{n=1}^{\infty} a_n$ and $\prod_{n=1}^{\infty} b_n$ converges, hence $\prod_{n=1}^{\infty} a_n=l\in[0,\infty)$ and $\prod_{n=1}^{\infty} b_n=m \in [0,\infty)$; so $$\left(\lim_{k\to\infty}\prod_{n=1}^{k} a_n\right)\cdot\left(\lim_{k \to \infty}\prod_{n=1}^{k} b_n\right)=lm$$ Hence the product converges to $lm$.
Are these solutions correct?