Question

Let $r$ be any real number and $(a_n) = \frac{1}{n^r}$ be sequence of real number, then}

• A. $(a_n)$ is always convergent for any $r \in R$
• B. $(a_n)$ is convergent only for $r = 1$
• C. $(a_n)$ is convergent for any $r \geq 0$
• D. $(a_n)$ is divergent for any $r \in R$

Hint:

Do not confuse the convergence of a sequence with the convergence of an infinite series:
- The sequence \(\left(\frac{1}{n^r}\right)\) converges for all \(\mathbf{r \ge 0}\).
- The series \(\sum \frac{1}{n^r}\) converges only for \(\mathbf{r > 1}\) (p-series test).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate the convergence of the sequence \((a_n) = \frac{1}{n^r}\) by calculating its limit as \(n \to \infty\) for different ranges of the exponent \(r\).

Step 3: Detailed Explanation:
Let us analyze the limit behavior of the sequence \(a_n = \frac{1}{n^r}\) across different values of \(r \in \mathbb{R}\):
1. Case 1: \(r > 0\):
As \(n \to \infty\), the term \(n^r\) grows infinitely large.
Therefore, the reciprocal approaches zero:
\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{n^r} = 0 \] Since the limit is a unique real number (0), the sequence converges for all \(r > 0\).
2. Case 2: \(r = 0\):
For any \(n \ge 1\), the term \(n^0 = 1\).
The sequence becomes a constant sequence:
\[ a_n = \frac{1}{n^0} = \frac{1}{1} = 1 \] This constant sequence converges to 1:
\[ \lim_{n\to\infty} a_n = 1 \] Therefore, the sequence converges when \(r = 0\).
3. Case 3: \(r < 0\):
Let \(r = -p\), where \(p > 0\). We can rewrite the general term as:
\[ a_n = \frac{1}{n^{-p}} = n^p \] As \(n \to \infty\), the term \(n^p\) grows infinitely large:
\[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} n^p = \infty \] Because the limit is infinite, the sequence is divergent for all \(r < 0\).
Combining these cases, the sequence \((a_n)\) is convergent if and only if \(r \ge 0\).

Step 4: Final Answer:
The sequence is convergent for any \(r \ge 0\), matching Option (C).