Question

Consider the sequences \(\{a_n\}\), \(\{b_n\}\), \(\{c_n\}\) defined for \(n \in \mathbb{N}\) by \[\text{(I) } a_n = n^2 \sin\left(\frac{1}{n}\right), \qquad \text{(II) } b_n = 1 + \frac{(-1)^n}{n}, \qquad \text{(III) } c_n = n\cos\left(\frac{1}{n}\right).\] Pick out the correct option.

• A. Only (II) is a Cauchy sequence.
• B. Only (I) and (II) are Cauchy sequences.
• C. Only (II) and (III) are Cauchy sequences.
• D. Only (I) and (III) are Cauchy sequences.

Hint:

A Cauchy sequence in R must converge; check whether each sequence is bounded and convergent.

Answer 1

The Correct Option is A

Step 1: Recall the criterion. A sequence of real numbers is Cauchy if and only if it converges, since \(\mathbb{R}\) is complete. So it suffices to check convergence of each sequence.

Step 2: Analyze \(a_n = n^2\sin(1/n)\). Using \(\sin(1/n) = \frac{1}{n} - \frac{1}{6n^3} + O(n^{-5})\), \[a_n = n^2\left(\frac1n - \frac{1}{6n^3}+\cdots\right) = n - \frac{1}{6n} + \cdots \to \infty.\] So \(\{a_n\}\) diverges and is not Cauchy.

Step 3: Analyze \(b_n = 1 + (-1)^n/n\). As \(n\to\infty\), \((-1)^n/n \to 0\), so \(b_n \to 1\). A convergent sequence is Cauchy, so \(\{b_n\}\) is Cauchy.

Step 4: Analyze \(c_n = n\cos(1/n)\). Using \(\cos(1/n) = 1 - \frac{1}{2n^2}+O(n^{-4})\), \[c_n = n\left(1 - \frac{1}{2n^2}+\cdots\right) = n - \frac{1}{2n}+\cdots \to \infty.\] So \(\{c_n\}\) diverges and is not Cauchy.

Step 5: Conclusion. Only \(b_n\) (statement II) is a Cauchy sequence.

\[\boxed{\text{Only (II) is Cauchy}}\]