Sequences & Series Question - None

Question

Statement-1: For every natural number $n \geq 2$, $\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} > \sqrt{n}$.

Statement-2: For every natural number $n \geq 2$, $\sqrt{n(n+1)} < n+1$.

Accepted Answer

Step 1: Verify Statement-1

For the sum S = 1/√1 + 1/√2 + ... + 1/√n, use integral comparison:

∫₁ⁿ 1/√x dx < S (lower Riemann sum)

This integral equals 2√n - 2

However, more precisely: S > ∫₁ⁿ 1/√x dx = 2(√n - 1)

For n ≥ 2: 2(√n - 1) ≥ √n requires 2√n - 2 ≥ √n, or √n ≥ 2, true for n ≥ 4

Direct verification for n = 2, 3 confirms S > √n. Statement-1 is TRUE

Step 2: Verify Statement-2

Need to prove: √(n(n+1)) < n + 1 for n ≥ 2

Square both sides: n(n+1) < (n+1)²

This simplifies: n(n+1) < n² + 2n + 1

Which gives: n² + n < n² + 2n + 1

Simplifying: 0 < n + 1 ✓ (always true)

Statement-2 is TRUE

Step 3: Determine Relationship

Statement-2 is a simple algebraic fact, while Statement-1 requires deeper analysis via integral bounds. Statement-2 does NOT provide reasoning for Statement-1.

∴ Both statements are true, but Statement-2 is NOT a correct explanation of Statement-1. Answer: B

Question:

Step-by-Step Solution

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