Using principle of mathematical induction prove that `sqrtn<1/sqrt1+1/sqrt2+1/sqrt3+......+1/sqrtn` for all natural numbers `n >= 2`.

Explanation

Consider the statement
`P(n):sqrt(n)lt(1)/sqrt(1)+(1)/sqrt(2)+ . . . +(1)/sqrt(n)`, for all natural numbers `nle2`. Step I We obser that P(2) is true.
`P(2):sqrt(2)lt(1)/sqrt(1)+(1)/sqrt(2),` which is true.
Step II Now, assume that P(n) is true for n=k.
`P(k):sqrt(k)lt(1)/sqrt(1)+(1)/sqrt(2)+ . . . +(1)/sqrt(k)`, is true.
Step III To prove P(k+1) is true, we have to show that
`P(k+1):sqrt(k+1)lt(1)/sqrt(1)+(1)/sqrt(2)+ . . . +(1)/sqrt(k+1)`, is true.
Given that, `sqrt(k)lt(1)/sqrt(1)+(1)/sqrt(2)+ . . . +(1)/sqrt(k)`,
`rArrsqrt(k)+(1)/sqrt(k+1)lt(1)/sqrt(1)+ . . . +(1)/sqrt(k)+(1)/(sqrt(k+1))` . . .(i)
`rArr((sqrt(k))(sqrt(k+1))+1)/(sqrt(k+1))(1)/sqrt(k+1)lt(1)/sqrt(1)+ . . . +(1)/sqrt(k)+(1)/(sqrt(k+1))`
if `sqrt(k+1)lt(sqrt(k)sqrt(k+1)+1)/(sqrt(k+1))`
`rArrk+1ltsqrt(k)sqrt(k+1)+1`
`rArrkltsqrt(k(k+1))rArrsqrt(k)ltsqrt(k)+1` . . .(ii)
From Eqs. (i) and (ii),
`sqrt(k+1)lt(1)/sqrt(1)+(1)/(sqrt(2))+. . .+(1)/sqrt(k+1)`
So, P(k+1) is true, whenever P(k) is true. Hence, P(n) is true.