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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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