English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Im not sure if you have to use induction or not, but I am pretty sure divisibility is involved.

2007-11-06 10:48:49 · 2 answers · asked by gregie09 1 in Science & Mathematics Mathematics

2 answers

Okay, add up all the fractions, which will get you:

(1/n!)( (2*3...n) + (1*3...n) + ....+ (1*2...(n-2)(n)) + (1*2...(n-1)) )

Notice that each term in the numerator is missing one number in the series 1 to n. Consider the largest prime number p < n, so that p appears only once in the series 1 to n. All but one of the terms will have that prime number. That means p in the denominator n! will divide into all of those terms except that one which is missing that p. Hence, the sum cannot be an integer.

2007-11-06 11:00:06 · answer #1 · answered by Scythian1950 7 · 6 1

Someone else just answered this exact question. Look a page or two down in the mathematics section.

2007-11-06 18:52:55 · answer #2 · answered by Sean H 5 · 0 0

fedest.com, questions and answers