Oh Really... Topic: S&S/Sets. Level: AIME/Olympiad.

Problem: (1999 IMC Day 1 - #2) Does there exist a bijective map $ f(n): \mathbb{N} \rightarrow \mathbb{N} $ so that $ \displaystyle \sum_{n=1}^{\infty} \frac{f(n)}{n^2} $ is finite?

Solution: Consider the bijective map $ f_k(n): \{1,2, \ldots, k\} \rightarrow \{1,2, \ldots k\} $ and the sum $ \displaystyle S_k = \sum_{n=1}^k \frac{f_k(n)}{n^2} $. Since the sets

$ \{1, 2, \ldots, k \} $ and $ \{\frac{1}{1^2}, \frac{1}{2^2}, \cdots, \frac{1}{k^2} \} $

are oppositely ordered, by Rearrangement we have

$ \displaystyle S_k \ge \frac{1}{1^2}+\frac{2}{2^2}+\cdots+\frac{k}{k^2} = \sum_{n=1}^k \frac{1}{n} $.

Hence $ \displaystyle \lim_{k \rightarrow \infty} S_k \ge \lim_{k \rightarrow \infty} \sum_{n=1}^k \frac{1}{n} = \infty $, so the answer is no. QED.

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Comment: Basically, knowledge of the Rearragement inequality killed this problem. It's also useful to know what a bijective map is - a function $ f:A \rightarrow B $ where $ A $ and $ B $ are sets such that for any $ x, y \in A $ we have $ f(x) = f(y) \Rightarrow x = y $ and for any $ b \in B $ there exists $ x \in A $ such that $ f(x) = b $. A lot of higher level math is vocabulary and notation, which is sort of a pity, but that's just how it is.

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Practice Problem: Evaluate $ \displaystyle \int_0^{\infty} e^{-x^2}dx $.