ICAR AIEEA PG 2024 - Statistical Sciences

Convergence analysis:

• For $(a_n) = \frac{n+1}{n} = 1 + \frac{1}{n}$:
  $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = 1 + 0 = 1$$
  Since the limit exists and is finite, $(a_n)$ converges to 1.
• For $(b_n) = (-1)^n$:
  The terms alternate: $-1, 1, -1, 1, \dots$. As $n \to \infty$, $(b_n)$ has two subsequential limit points (-1 and +1), so $\lim_{n\to\infty} (-1)^n$ does not exist. Hence $(b_n)$ is oscillatory (not convergent).

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.