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

Explanation

Given,

`sqrtn<1/sqrt1+1/sqrt2+1/sqrt3+......+1/sqrtn`

where p(k) is true

let p(k+1) is also true,

`=>sqrt(k+1)<1/sqrt1+1/sqrt2+1/sqrt3+......+1/(sqrtn(k+1))`

`=>sqrt(k+1)+1/(sqrt(k+1))<1/sqrt1+1/sqrt2+1/sqrt3+......+1/(sqrtn(k+1))+1/(sqrt(k))`

So,

`sqrt(k+1)<(sqrt(k)sqrt(k+1+1))/sqrt(k+1)`

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