PUMaC 2011 · 数论（A 组） · 第 5 题
PUMaC 2011 — Number Theory (Division A) — Problem 5
题目详情
- [ 5 ] Let d ( n ) denote the number of divisors of n (including itself). You are given that ∞ 2 ∑ 1 π = . 2 n 6 n =1 Find p (6), where p ( x ) is the unique polynomial with rational coefficients satisfying ∞ ∑ d ( n ) p ( π ) = . 2 n n =1 3 4
解析
- First, show that this sum converges by showing that d ( n ) ≤ 2 n for all n ≥ 1. Let x be some √ √ divisor of n with x ≤ n . There is a one to one correspondence between divisors at most n √ √ n and divisors at least n (map x to for any x ≤ n ). However, there are at most n divisors x √ √ of n that are at most n , so d ( n ) ≤ 2 n for all positive integers n . Therefore, the given sum is bounded above by ∞ ∑ 2 3 2 x n =1 which converges. ∑ ∑ d ( n ) 1 First solution : Note that d ( n ) = 1, so = . This means that the desired 2 2 2 n k ( n/k ) k | n k | n sum is equal to ∞ ∞ ∞ ∑ ∑ ∑ ∑ 1 1 = . 2 2 2 2 k ( n/k ) k m n =1 k =1 m =1 k | n n This can be seen by letting m = and switching the order of summation (which is valid by k the absolute convergence of the original series). Therefore, we can write this series as ( ) 2 ∞ 4 ∑ 1 π = . 2 m 36 m =1 4 x Hence p ( x ) = and p (6) = 36 . p ( x ) is unique because π is transcendental. 36 Second solution : Since the original sum converges, one may use the unique prime factoriza- tion of the integers and the fact that d ( n ) is multiplicative to factor the given sum as ∞ ∞ j ∏ ∑ d ( p ) i 2 j p i i =1 j =0 j where p is the i th prime. Note also that d ( p ) = j + 1 for any prime p . Now, compute the i sum ∞ ∑ j S ( x ) = ( j + 1) x j =0 ∑ ∑ ∞ ∞ j j To compute a closed form for S ( x ), consider xS ( x ) = jx and S ( x ) − xS ( x ) = x = j =1 j =0 ( ) 2 ∑ ∞ 1 1 1 j for | x | < 1. Therefore, S ( x ) = = x . Letting x = for each term of the 2 j =0 1 − x (1 − x ) p i 2 ∑ 4 4 ∞ 1 π x 2 infinite product shows that the desired sum is equal to ( ) = . Hence p ( x ) = 2 j =1 j 36 36 and p (6) = 36 . 3 4