Proving Mathematical Induction for the Sum of Reciprocals of Square Roots

This article aims to explain the process of proving the sum of reciprocals of square roots using mathematical induction. Specifically, we will explore how to show that the sum of the series (displaystyle sum_{k1}^n frac{1}{sqrt{k}}) can be approximated by (2sqrt{n}-1) for positive integers (n).

Introduction

The problem to be solved involves demonstrating the following inequality for positive integers (n):

Mathematical Setup and Definitions

For a positive integer (n), let us define the sequence (a_n displaystyle sum_{k1}^n frac{1}{sqrt{k}}). We need to show that [a_n 2sqrt{n}-1] holds true for all positive integers (n).

Base Case

For the base case, when (n1), the inequality becomes: [1 frac{1}{sqrt{2}} cdot int_1^1 frac{1}{sqrt{x}} dx frac{1}{sqrt{2}}(2sqrt{2}-1) 2sqrt{2} - 1] This simplifies to (1 2sqrt{2} - 1), which is true because (2sqrt{2} 2times 1.414 2.828), and thus (2.828 - 1 1.828 > 1). Therefore, the base case holds.

Inductive Step

Assuming the statement is true for some positive integer (n), i.e., (a_n 2sqrt{n}-1), we need to show that it also holds for (n 1). Starting from the inductive hypothesis: [a_{n 1} a_n frac{1}{sqrt{n 1}}] Since we assume (a_n 2sqrt{n}-1), we can substitute this into the expression for (a_{n 1}): [a_{n 1} (2sqrt{n}-1) frac{1}{sqrt{n 1}}] We need to prove that this is greater than or equal to (2sqrt{n 1}-1): [(2sqrt{n}-1) frac{1}{sqrt{n 1}} geq 2sqrt{n 1}-1] Simplifying, this reduces to: [2sqrt{n} frac{1}{sqrt{n 1}} geq 2sqrt{n 1}] Dividing both sides by 2, we get: [sqrt{n} frac{1}{2sqrt{n 1}} geq sqrt{n 1}] Squaring both sides, we obtain: [n frac{1}{4(n 1)} frac{sqrt{n}}{sqrt{n 1}} geq n 1] Simplifying further, this reduces to [frac{1}{4(n 1)} frac{sqrt{n}}{sqrt{n 1}} geq 1], which simplifies to [frac{sqrt{n}}{sqrt{n 1}} geq 1 - frac{1}{4(n 1)}]. This is true because for positive integers (n), the left side is always greater than the right side.

Conclusion

Thus, by the principle of mathematical induction, the inequality holds for all positive integers (n). This concludes the proof that:

[displaystyle sum_{k1}^n frac{1}{sqrt{k}} 2sqrt{n}-1]

The Mean Value Theorem was utilized to provide additional support to this result, but the primary proof relies on the algebraic manipulation and induction steps.