Statements regarding sequences and series

If $(a_n), (b_n)$ are sequences then

$1.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = L>0$ then $\lim_{n\rightarrow\infty} a_nb_n = \infty$. True

$2.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = -\infty$ then $\lim_{n\rightarrow\infty} (a_n+b_n) = \infty$. False since $\infty-\infty$ is undefined.

$3.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = -\infty$ then $\lim_{n\rightarrow\infty} a_nb_n = -\infty$. True

$4.$ If neither $(a_n), (b_n)$ converge then $(a_nb_n)$ doesn't converge. False since $(\frac{1}{n})$ and $n$ doesn't converge individually but when multplied they converge to 1.

$5.$ If $(|a_n|)$ converge then $(a_n)$ converge. False since $(|(-1)^n|)$ converge but $((-1)^n)$ doesnt converge. However the converse of this statement is true.

$6.$ If $a_n=0$ for every $n$ then $\sum a_n$ converges. True. Sum converges to $0$.

$7.$ If $\sum a_n$ converges then $\sum \frac{1}{a_n}$ diverges to infinity. True

$8.$ If $\sum a_n$, $\sum b_n$ converges then $\sum (a_n + b_n)$ converges. True

$9.$ If $a_n>c>0$ for every $n$ then $\sum a_n$ diverges to infinity. True

$10.$ If $\sum a_n$ diverges and $(b_n)$ is bounded then $\sum a_nb_n$ diverges to infinity. True

$11.$ If $a_n>0$ and $\sum a_n$ converges then $\sum (a_n)^2$ converges. True.

Are these answers correct?

Answer

Most of your answers look alright. A few remarks:

$7.$ If $\sum a_n$ converges then $\sum \frac{1}{a_n}$ diverges to infinity. True

Do you know anything about the sign of $a_n$? If not, note that the convergence of $\sum a_n$ implies that $a_n \to 0$, but not that $a_n > 0$; i.e. it's possible that $a_n <0$.

Also note that it doesn't even have to diverge to $\pm\infty$, for example:
$$a_n = \frac{(-1)^n}{n} \implies \frac{1}{a_n} = (-1)^nn$$

$10.$ If $\sum a_n$ diverges and $(b_n)$ is bounded then $\sum a_nb_n$ diverges to infinity. True

If $\sum a_n$ diverges, it doesn't necessarily diverge to infinity (think of $1-1+1-1+1-\ldots$ for example); now take $b_n \equiv 1$.

And as pointed out by Akiva Weinberger in the comment, you can even make it converge (e.g. take $b_n = \tfrac{1}{n}$ and keep $a_n$ as above).