HMMT 二月 2008 · 代数 · 第 10 题
HMMT February 2008 — Algebra — Problem 10
题目详情
- [ 8 ] Evaluate the infinite sum ( ) ∞ ∑ 2 n 1 . n n 5 n =0 1
解析
- [ 8 ] Evaluate the infinite sum ( ) ∞ ∑ 2 n 1 . n n 5 n =0 √ Answer: 5 First Solution: Note that ( ) 2 n (2 n )! (2 n )(2 n − 2)(2 n − 4) · · · (2) (2 n − 1)(2 n − 3)(2 n − 5) · · · (1) = = · n n ! · n ! n ! n ! ( ) ( ) ( ) ( ) n ( − 2) 1 1 1 1 n = 2 · − − − 1 − − 2 · · · − − n + 1 n ! 2 2 2 2 ( ) 1 − n 2 = ( − 4) . n 1 Then, by the binomial theorem, for any real x with | x | < , we have 4 ( ) ( ) ∞ ∞ 1 ∑ ∑ − 2 n − 1 / 2 n n 2 (1 − 4 x ) = ( − 4 x ) = x . n n n =0 n =0 Therefore, ( ) ( ) ∞ n ∑ √ 2 n 1 1 = √ = 5 . n 5 4 1 − n =0 5 Second Solution: Consider the generating function ( ) ∞ ∑ 2 n n f ( x ) = x . n n =0 It has formal integral given by ( ) ∞ ∞ ∞ ∑ ∑ ∑ 1 2 n n +1 n +1 n g ( x ) = I ( f ( x )) = x = C x = x C x , n n n + 1 n n =0 n =0 n =0 ( ) ∑ ∞ 2 n 1 n where C = is the n th Catalan number. Let h ( x ) = C x ; it suffices to compute this n n n =0 n +1 n generating function. Note that ( ) k ∑ ∑ ∑ ∑ 2 i + j k k +1 1 + xh ( x ) = 1 + x C C x = 1 + x C C x = 1 + C x = h ( x ) , i j i k − i k +1 i,j ≥ 0 i =0 k ≥ 0 k ≥ 0 where we’ve used the recurrence relation for the Catalan numbers. We now solve for h ( x ) with the quadratic equation to obtain √ √ 2 1 /x ± 1 /x − 4 /x 1 ± 1 − 4 x h ( x ) = = . 2 2 x 1 Note that we must choose the − sign in the ± , since the + would lead to a leading term of for h (by x √ expanding 1 − 4 x into a power series). Therefore, we see that √ ( ) 1 − 1 − 4 x 1 f ( x ) = D ( g ( x )) = D ( xh ( x )) = D = √ 2 1 − 4 x √ and our answer is hence f (1 / 5) = 5. 3