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Question
Prove the statement by using the Principle of Mathematical Induction:
`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.
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Solution
Let P(n): `sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, ∀ n ≥ 2
Step 1: P(2): `sqrt(2) < 1/sqrt(1) + 1/sqrt(2)` which is true.
Step 2: P(k): `sqrt(k) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)`
Let it be true.
Step 3: P(k + 1): `sqrt(k + 1) < 1/sqrt(2) + 1/sqrt(2) + ... + 1/sqrt(k + 1)`
From Step 2, we have
`sqrt(k) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k)`
⇒ `sqrt(k) + 1/sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) + 1/sqrt(k + 1)`
⇒ `(sqrt(k) . sqrt(k + 1) + 1)/sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k) + 1/sqrt(k + 1)` ......(i)
Now if `sqrt(k + 1) < (sqrt(k) . sqrt(k + 1) + 1)/sqrt(k + 1)`
⇒ `(k + 1) < sqrt(k) . sqrt(k + 1) + 1`
⇒ `k < sqrt(k) . sqrt(k + 1)` .......(ii)
From equation (i) and (ii) we get
`sqrt(k + 1) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(k + 1)`
Hence, P(k + 1) is true whenever P(k) is true.
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