Key Concepts and Formulas

• Inequalities: Comparing mathematical expressions and manipulating them while preserving the inequality.
• Telescoping Series: A series where most terms cancel out, leaving only a few terms.
• Mathematical Induction: A method of proving a statement for all natural numbers by showing it holds for a base case and that if it holds for $n$, it also holds for $n+1$.

Step-by-Step Solution

• Step 1: Analyze Statement 2

  • What & Why: We want to determine if Statement 2, $\sqrt{n(n+1)} < n+1$, is true for all $n \ge 2$. We will manipulate the inequality to see if it holds.
  • Math: Squaring both sides (since both sides are positive), we get: $(\sqrt{n(n+1)})^2 < (n+1)^2$ $n(n+1) < (n+1)^2$ $n^2 + n < n^2 + 2n + 1$
  • Reasoning: Squaring preserves the inequality because both sides are positive. Expanding the terms allows us to simplify.

• Step 2: Simplify the Inequality in Statement 2

  • What & Why: Continue simplifying the inequality obtained in Step 1 to arrive at a simpler, more obvious inequality.
  • Math: Subtracting $n^2 + n$ from both sides: $0 < n + 1$
  • Reasoning: This simplification isolates the variable and reveals the underlying relationship.

• Step 3: Conclude on Statement 2

  • What & Why: Based on the simplified inequality, determine the truthfulness of Statement 2.
  • Reasoning: Since $0 < n + 1$ is true for all natural numbers $n$, it is true for all $n \ge 2$. Therefore, Statement 2 is true.

• Step 4: Analyze Statement 1

  • What & Why: Determine if Statement 1, $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} > \sqrt{n}$, is true for all $n \ge 2$.
  • Reasoning: We will attempt to prove this using a comparison.

• Step 5: Manipulate the inequality $\frac{1}{\sqrt{k}}$

  • What & Why: We aim to find a useful upper bound for $\frac{1}{\sqrt{k}}$. Consider the expression $2(\sqrt{k} - \sqrt{k-1})$.
  • Math: $2(\sqrt{k} - \sqrt{k-1}) = 2\frac{(\sqrt{k} - \sqrt{k-1})(\sqrt{k} + \sqrt{k-1})}{\sqrt{k} + \sqrt{k-1}} = \frac{2(k - (k-1))}{\sqrt{k} + \sqrt{k-1}} = \frac{2}{\sqrt{k} + \sqrt{k-1}}$
  • Reasoning: Multiplying by the conjugate helps rationalize the numerator.

• Step 6: Compare $\frac{1}{\sqrt{k}}$ with the Manipulated Inequality

  • What & Why: Relate $\frac{1}{\sqrt{k}}$ to the expression derived in the previous step.
  • Math: Since $\sqrt{k} > \sqrt{k-1}$ for $k > 1$, then $\sqrt{k} + \sqrt{k-1} < \sqrt{k} + \sqrt{k} = 2\sqrt{k}$. Therefore, $\frac{2}{\sqrt{k} + \sqrt{k-1}} > \frac{2}{2\sqrt{k}} = \frac{1}{\sqrt{k}}$. Thus, $\frac{1}{\sqrt{k}} < 2(\sqrt{k} - \sqrt{k-1})$.
  • Reasoning: We found an inequality relating the term in the sum to a difference of square roots.

• Step 7: Sum the Inequality from Step 6

  • What & Why: Sum the inequality from $k=1$ to $n$ to see if it leads to a contradiction or confirms the statement.
  • Math: $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} < \sum_{k=1}^{n} 2(\sqrt{k} - \sqrt{k-1}) = 2\sum_{k=1}^{n} (\sqrt{k} - \sqrt{k-1})$
  • Reasoning: Summing the inequality term by term preserves the inequality.

• Step 8: Evaluate the Telescoping Sum

  • What & Why: Evaluate the sum on the right side of the inequality in Step 7.
  • Math: The sum is a telescoping sum: $2[(\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + \dots + (\sqrt{n} - \sqrt{n-1})] = 2(\sqrt{n} - \sqrt{0}) = 2\sqrt{n}$ Thus, $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} < 2\sqrt{n}$.

• Step 9: Refine the Summation and Inequality

  • What & Why: We want a tighter upper bound. We have $\frac{1}{\sqrt{k}} < 2(\sqrt{k} - \sqrt{k-1})$ for $k>1$. Thus, $\sum_{k=2}^{n} \frac{1}{\sqrt{k}} < 2 \sum_{k=2}^n (\sqrt{k} - \sqrt{k-1}) = 2(\sqrt{n} - 1)$ Adding $1/\sqrt{1} = 1$ to both sides, $\sum_{k=1}^n \frac{1}{\sqrt{k}} < 1 + 2(\sqrt{n} - 1) = 2\sqrt{n} - 1$
  • Reasoning: This gives a more precise upper bound for the sum. This shows that Statement 1, $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} > \sqrt{n}$, is not necessarily true. We have $\sum_{k=1}^n \frac{1}{\sqrt{k}} < 2\sqrt{n} - 1$.

• Step 10: Determine the Truthfulness of Statement 1

  • What & Why: Based on the inequalities derived, determine if Statement 1 is true or false.
  • Reasoning: We have $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} < 2\sqrt{n} - 1$. Since $\sum_{k=1}^{n} \frac{1}{\sqrt{k}}$ is less than $2\sqrt{n} - 1$, it is not necessarily greater than $\sqrt{n}$. Therefore, Statement 1 is FALSE.

• Step 11: Determine if Statement 2 Explains Statement 1

  • What & Why: Determine if the truth of Statement 2 explains the falsehood of Statement 1.
  • Reasoning: Statement 2 is about the relationship between $\sqrt{n(n+1)}$ and $n+1$, while Statement 1 is about a sum of reciprocals of square roots. The truth of Statement 2 does not provide any insight into why Statement 1 is false. Therefore, Statement 2 does not explain Statement 1.

• Step 12: Conclude

  • What & Why: State the truthfulness of both statements and the relationship between them.
  • Reasoning: Statement 1 is FALSE, and Statement 2 is TRUE. Statement 2 does not explain Statement 1.

Common Mistakes & Tips

• When dealing with inequalities, remember to consider the signs of the terms involved before performing operations like squaring.
• Telescoping series are a powerful tool for simplifying sums, but remember to correctly identify the terms that cancel out.
• Mathematical induction requires a solid base case and a valid inductive step. Ensure that the inductive step holds for all $n$ greater than or equal to the base case.

Summary Statement 2, $\sqrt{n(n+1)} < n+1$, is true because it simplifies to $0 < n + 1$, which holds for all $n \ge 2$. Statement 1, $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} > \sqrt{n}$, is false. We showed that $\sum_{k=1}^n \frac{1}{\sqrt{k}} < 2\sqrt{n} - 1$. Statement 2 does not explain Statement 1. Therefore, the correct answer is (A).

Final Answer The final answer is \boxed{A}, which corresponds to option (A).