Statement-1: For every natural number `n ge 2, (1)/(sqrt1)+(1)/(sqrt2)+…..(1)/(sqrtn) gt sqrtn`
Statement-2: For every natural number `n ge 2, sqrt(n(n+1) lt n+1`

Answer

To solve the given problem, we need to analyze both statements and prove their validity step by step.

### Step 1: Analyze Statement 1
**Statement 1:** For every natural number \( n \geq 2 \),
\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{n}} > \sqrt{n} \]

**Proof:**
1. **Base Case:** Let's check for \( n = 2 \):
\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{\sqrt{2}} > \sqrt{2} \]
Since \( 1 + \frac{1}{\sqrt{2}} \approx 1.707 > 1.414 \), the base case holds.

2. **Inductive Step:** Assume the statement is true for \( n = k \):
\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} > \sqrt{k} \]
We need to prove it for \( n = k + 1 \):
\[ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \ldots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} > \sqrt{k + 1} \]
By the inductive hypothesis:
\[ > \sqrt{k} + \frac{1}{\sqrt{k+1}} \]
We need to show:
\[ \sqrt{k} + \frac{1}{\sqrt{k+1}} > \sqrt{k + 1} \]
Squaring both sides:
\[ k + 2\frac{\sqrt{k}}{\sqrt{k+1}} + \frac{1}{k+1} > k + 1 \]
Simplifying gives:
\[ 2\frac{\sqrt{k}}{\sqrt{k+1}} + \frac{1}{k+1} > 1 \]
This inequality holds true for \( k \geq 2 \). Thus, by induction, Statement 1 is true for all \( n \geq 2 \).

### Step 2: Analyze Statement 2
**Statement 2:** For every natural number \( n \geq 2 \),
\[ \sqrt{n(n+1)} < n + 1 \]

**Proof:**
1. **Base Case:** Check for \( n = 2 \):
\[ \sqrt{2 \cdot 3} = \sqrt{6} < 3 \]
This is true since \( \sqrt{6} \approx 2.45 < 3 \).

2. **Inductive Step:** Assume it holds for \( n = k \):
\[ \sqrt{k(k+1)} < k + 1 \]
We need to prove it for \( n = k + 1 \):
\[ \sqrt{(k + 1)(k + 2)} < k + 2 \]
Squaring both sides:
\[ (k + 1)(k + 2) < (k + 2)^2 \]
Expanding both sides:
\[ k^2 + 3k + 2 < k^2 + 4k + 4 \]
Simplifying gives:
\[ 2 < k + 2 \]
This is true for \( k \geq 2 \). Thus, by induction, Statement 2 is true for all \( n \geq 2 \).

### Conclusion
Both statements are true for every natural number \( n \geq 2 \).

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