Question

С помощью метода математической индукции решить 1/1!+1/2!+1/3!+...+1/n!<2n-1/n, n>3

Answer 1

1) n = 4
1/1! + 1/2! + 1/3! + 1/4! = 1 + 1/2 + 1/6 + 1/24 = 41/24 < (8-1)/4 = 7/4 = 42/24 - верно

2) n = k
1/1! + 1/2! + ... + 1/k! < (2k-1)/k

3) n = k + 1
1/1! + 1/2! + ... + 1/k! + 1/(k+1)! < (2k-1)/k + 1/(k(k+1)) = ((2k-1)(k+1)+1)/(k(k+1)) = (2k² - k + 2k - 1 + 1)/(k(k+1)) = k(2k+1)/(k(k+1)) = (2k+1)/(k+1) 

значит:

1/1! + 1/2! + ... + 1/k! + 1/(k+1)! < (2k+1)/(k+1) 

по ММИ доказано.