Finite or Not

Calculus Level 3

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

Yes No

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
May 14, 2019

Suppose that f : N N f \,:\,\mathbb{N} \to \mathbb{N} is injective. For any N N N \in \mathbb{N} , suppose that { f ( 1 ) , f ( 2 ) , . . . , f ( N ) } = { a 1 , a 2 , . . . , a N } \{f(1),f(2),...,f(N)\} = \{a_1,a_2,...,a_N\} , where a 1 < a 2 < < a N a_1 < a_2 < \cdots < a_N . By the Rearrangement Lemma, n = 1 N f ( n ) n 2 n = 1 N a n n 2 \sum_{n=1}^N \frac{f(n)}{n^2} \; \ge \; \sum_{n=1}^N \frac{a_n}{n^2} Since a 1 < a 2 < < a N a_1 < a_2 < \cdots < a_N , we deduce that a n n a_n \ge n for 1 n N 1 \le n \le N , and hence n = 1 N f ( n ) n 2 n = 1 N a n n 2 n = 1 N n n 2 \sum_{n=1}^N \frac{f(n)}{n^2} \; \ge \; \sum_{n=1}^N \frac{a_n}{n^2} \; \ge \;\sum_{n=1}^N \frac{n}{n^2} and hence n = 1 N f ( n ) n 2 > ln N N N \sum_{n=1}^N \frac{f(n)}{n^2} \; > \; \ln N \hspace{2cm} N \in \mathbb{N} which means that the infinite series n = 1 f ( n ) n 2 \sum_{n=1}^\infty \frac{f(n)}{n^2} diverges.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...