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Question
Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.
Solution
Let P(n): `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24` ∀ n ∈ N
Step 1: P(2) : `1/(2 + 1) + 1/(2 + 2) > 13/24`
⇒ `1/3 + 1/4 > 13/24`
⇒ `7/12 > 13/24`
⇒ `14/24 > 13/24` which is true for P(2).
Step 2: P(k) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`.
Let it be true for P(k).
Step 3: P(k + 1) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`
Since `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`
So `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.
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