1·3·5·...·(2k-1)/(2·4·6·...·(2k))
= 1/2·3/4·5/6·...·(2k-1)/(2k)
< 2/3·4/5·5/7·...·(2k)/(2k+1)
= (2·4·6·...·(2k))/(3·5·7·...·(2k+1)).
故(1·3·5·...·(2k-1))²/(2·4·6·...·(2k))²
< (1·3·5·...·(2k-1))/(2·4·6·...·(2k))·(2·4·6·...·(2k))/(3·5·7·...·(2k+1))
= (1·3·5·...·(2k-1))/(3·5·7·...·(2k+1))
= 1/(2k+1).
即得1·3·5·...·(2k-1)/(2·4·6·...·(2k)) < 1/√(2k+1).
而1/√(2k+1) = 2/(√(2k+1)+√(2k+1))
< 2/(√(2k+1)+√(2k-1))
= 2(√(2k+1)-√(2k-1))/((2k+1)-(2k-1))
= √(2k+1)-√(2k-1).
于是1·3·5·...·(2k-1)/(2·4·6·...·(2k)) < √(2k+1)-√(2k-1).
對k = 1,2,...,n求和即得:
1/2+1·3/(2·4)+1·3·5/(2·4·6)+...+1·3·5·...·(2n-1)/(2·4·6·...·(2n))
< (√3-√1)+(√5-√3)+(√7-√5)+...+(√(2n+1)-√(2n-1))
= √(2n+1)-1.