Evaluation of limit of algebraic sequence:

• $1$ neglects the factor of $2$ in the AP sum formula.
• The sum of the first $n$ natural numbers is given by $\sum_{k=1}^n k = \frac{n(n + 1)}{2}$. Substituting into the expression: $\lim_{n \to \infty} \frac{n(n + 1)}{2 n^2} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1 + 0}{2} = \frac{1}{2}$.
• $0$ would occur if the denominator had higher degree.
• $2$ is the reciprocal.

Application of the Squeeze (Sandwich) Theorem:

• Since $-1 \le \sin n \le 1$ for all natural numbers $n$, dividing by positive $n$ gives: $-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n}$. As $n \to \infty$, both $-\frac{1}{n} \to 0$ and $\frac{1}{n} \to 0$. By the Squeeze Theorem, $\lim_{n \to \infty} \frac{\sin n}{n} = 0$.
• $\lim_{x \to 0} \frac{\sin x}{x} = 1$, but here $n \to \infty$, not $0$.
• The limit exists and is unique ($0$).
• $1/2$ is incorrect.

Analysis of p-series convergence:

• Both series are convergent, so neither is divergent.
• For $\sum \frac{1}{n^2}$, the exponent is $p = 2 \gt 1$, which converges by the p-series test. For $\sum \frac{1}{n^3}$, the exponent is $p = 3 \gt 1$, which also converges by the p-series test. Therefore, both series are convergent.
• Both are convergent.
• Both are convergent.