prove (\frac{1}{2})(\frac{3}{4})(\frac{5}{6})..........(\frac{999999}{1000000}) < \frac{1}{1000}
1well its very easy ..
1/2<2/3
3/4<4/5
As n/n+1<n+1/n+2
multiplying we get n2+2n<n2+2n+1 =>0<1 which is true
similarly
999999/1000000<1000000/1000001
multiplying the above inequalities
1/2*3/4*4/5*...999999/1000000<2*4/3*..1000000/999999*(1/1000001)
let the quatity on the left side be p
then we get
p<(1/p)*(1/1000001)
thus
p<√(1/1000001)<√(1/1000000)<1/1000
1@rohan 200 your solution is perfect.
i thought up another solution
\frac{n-1}{n} < (\sqrt{\frac{n-1}{n}})(\sqrt{\frac{n}{n+1}}) is true for n>1 (it is easy to check by cross multiplication)
n=2 we have
\frac{1}{2}< (\sqrt{\frac{1}{2}})(\sqrt{\frac{2}{3}})
n=4 we have
\frac{3}{4}< (\sqrt{\frac{3}{4}})(\sqrt{\frac{4}{5}})
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.
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n=1000000 we have
\frac{999999}{1000000}< (\sqrt{\frac{999999}{1000000}})(\sqrt{\frac{1000000}{1000001}})
multiplyng the LHS and RHS of the above inequalities we get
(\frac{1}{2})(\frac{3}{4})(\frac{5}{6}).......(\frac{999999}{1000000})<\sqrt{\frac{1}{1000001}}<\frac{1}{1000} (as successive numerators and denominators get cancelled exept last one)
and hence the inequality