Q15. Find number of bijection from N to N such that

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.

Answer 1

Given that $f: \mathbb{N} \to \mathbb{N}$ is a bijective map and $\sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$.

 Note that, since $f$ is a injective function, $f(n)\geq  n$. It follows that

$$\frac{f(n)}{n^2}> \frac{n}{n^2}=\frac{1}{n}$$

By comparison test, $\sum_{n=1}^{\infty} \frac{f(n)}{n^2}$ is always divergent.

Hence, the correct answer is:

(B) zero.