Convergence of the sequence $(a_n) = \frac{1}{n^r}$:
If $r \gt 0$: $\lim_{n \to \infty} \frac{1}{n^r} = 0$, so $(a_n)$ converges to 0.
If $r = 0$: $a_n = \frac{1}{n^0} = 1$ for all $n$, so $(a_n)$ is a constant sequence converging to 1.
If $r \lt 0$: Let $r = -k$ where $k \gt 0$. Then $a_n = n^k \to \infty$ as $n \to \infty$, which diverges.

Fundamental theorems on real sequences:
Theorem: Every Cauchy sequence in $\mathbb{R}$ is bounded. (Option 4 is true).
Counter-example for Option 1 & 2: The sequence $a_n = (-1)^n$ is bounded ($-1 \le a_n \le 1$), but it is neither convergent nor a Cauchy sequence.
Option 3 is false: Because every Cauchy sequence is unconditionally bounded.

Consider the counterexample:
Let $a_n = (-1)^n$ and $b_n = (-1)^n$.
Both $(a_n)$ and $(b_n)$ are oscillatory and therefore divergent.
However, their quotient $\frac{a_n}{b_n} = \frac{(-1)^n}{(-1)^n} = 1$ is a constant sequence converging to 1.
Their product $a_n b_n = (-1)^n \times (-1)^n = (-1)^{2n} = 1$ is also a constant sequence converging to 1.

A series $\sum a_n$ is said to be absolutely convergent if the series of absolute values $\sum |a_n|$ converges.

Taking the absolute values of the terms:

By the p-series test, $\sum_{n=1}^\infty \frac{1}{n^p}$ converges whenever $p \gt 1$. Here $p = 2 \gt 1$, so the series of absolute values converges.