Question

Let \( f : \mathbb{N} \to \mathbb{N} \) be a bijective map such that \[ \sum_{n=1}^{\infty} \frac{f(n)}{n^2}<+\infty. \] The number of such bijective maps is

• A. exactly one.
• B. zero.
• C. finite but more than one.
• D. infinite.

Hint:

When a bijective function \(f:\mathbb{N}\to\mathbb{N}\) is involved in a convergent series, the only possible arrangement preserving convergence is the identity permutation.

Answer 1

The Correct Option is B

Step 1: Analyze the condition.
The series \(\sum_{n=1}^{\infty} \frac{f(n)}{n^2}\) must converge. Since \(f(n)\) is a bijection on \(\mathbb{N}\), it is a rearrangement of the natural numbers.
Step 2: Comparison with known series.
We know \(\sum_{n=1}^{\infty} \frac{n}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n}\), which diverges. Hence, for convergence, \(f(n)\) must not grow linearly or faster.
Step 3: Necessary condition.
For convergence, we must have \(f(n) \leq C\) for large \(n\), which is impossible for a bijection unless \(f(n)=n\). Thus, the only possibility is \(f(n)=n\) for all \(n\).
Step 4: Conclusion.
Hence, exactly one bijective map satisfies the condition, namely the identity function.