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

Explanation

Let `P(n) : sqrt(n) lt 1/(sqrt(1)) + (1)/(sqrt(2)) + "……" + (1)/(sqrt(n))`, for all natural numbers,
For ` n = 2, sqrt(2) lt (1)/(sqrt(1)) + (1)/(sqrt(2))` , which is true.
So, `P(2)` is true.
Let `P(n)` be true for some `n = k`.
Then `sqrt(k) lt (1)/(sqrt(1)) + (1)/(sqrt(2)) + "....." + (1)/(sqrt(k)) "......"(1)`
Adding `(1)/(sqrt(k+1))` on both sides, we get
`sqrt(k) + (1)/(sqrt(k+1)) lt (1)/(sqrt(1)) + (1)/(sqrt(2)) + "......" + (1)/(sqrt(k)) + (1)/(sqrt(k+1))"....."(2)`
Now, `sqrt(k+1) lt sqrt(k) + (1)/(sqrt(k+1))"......"(3)`
or `k 1 lt sqrt(k) sqrt(k+1) + 1`
or `k lt sqrt(k^(2) + k)`, where is true.
or `k + 1 lt sqrt(k) sqrt(k+1)+1`
or, ` k lt sqrt(k^(2) +k)`, which is true.
So from (2) and (3).
`sqrt(k+1)lt (1)/(sqrt(11)) + (1)/(sqrt(2)) + "......" + (1)/(sqrt(2)) + (1)/(sqrt(k+1))`
Thus, `P(k+1)` is true whenever `P(k)` is true.
So, by the principle of mathematical induction, `P(n)` is true for any natural number n.