Proving the Convergence of ( n^{sqrt{frac{1}{n}}} ) to 1: A Real Analysis Explained

In the realm of mathematics, real analysis plays a crucial role in understanding sequences and their behaviors as ( n ) approaches infinity. This article will delve into the process of proving that the sequence ( n^{sqrt{frac{1}{n}}} ) converges to 1. We will explore the steps involved, including the application of L'H?pital's Rule, and provide a detailed explanation for clarity and comprehension.

Introduction to the Problem

Consider the sequence defined by ( a_n n^{sqrt{frac{1}{n}}} ). Our goal is to prove that this sequence converges to 1 as ( n ) approaches infinity. To achieve this, we will follow a series of logical steps and mathematical transformations, ultimately arriving at a conclusive limit.

Step 1: Natural Logarithm Transformation

To tackle the problem, we begin by taking the natural logarithm of the sequence. Let's define ( y_n ) as the natural logarithm of ( a_n ): [ y_n ln left( n^{sqrt{frac{1}{n}}} right) ]

Using the properties of logarithms, we can simplify ( y_n ) as follows: [ y_n sqrt{frac{1}{n}} ln (n) ]

Step 2: Evaluating the Limit of ( y_n )

Next, we need to evaluate the limit of ( y_n ) as ( n ) approaches infinity. This can be written as:

[ lim_{n to infty} y_n lim_{n to infty} sqrt{frac{1}{n}} ln (n) ]

As ( n ) approaches infinity, ( sqrt{frac{1}{n}} ) approaches 0, while ( ln(n) ) approaches infinity. Therefore, the expression ( sqrt{frac{1}{n}} ln(n) ) is of the indeterminate form ( 0 cdot infty ). To resolve this indeterminacy, we can rewrite the expression in a form that allows us to apply L'H?pital's Rule:

[ lim_{n to infty} sqrt{frac{1}{n}} ln (n) lim_{n to infty} frac{ln n}{sqrt{n}} ]

Step 3: Applying L'H?pital's Rule

L'H?pital's Rule is applicable when we have the indeterminate forms ( frac{0}{0} ) or ( frac{infty}{infty} ). In our case, as ( n ) approaches infinity, both ( ln(n) ) and ( sqrt{n} ) approach infinity, making the form ( frac{infty}{infty} ) valid for L'H?pital's Rule.

Applying L'H?pital's Rule, we differentiate the numerator and the denominator with respect to ( n ): [ lim_{n to infty} frac{ln n}{sqrt{n}} lim_{n to infty} frac{frac{1}{n}}{frac{1}{2sqrt{n}}} ]

Further simplifying, we get: [ lim_{n to infty} frac{frac{1}{n}}{frac{1}{2sqrt{n}}} lim_{n to infty} frac{2sqrt{n}}{n} lim_{n to infty} frac{2}{sqrt{n}} ]

As ( n ) approaches infinity, ( frac{2}{sqrt{n}} ) approaches 0. Therefore, we have:

[ lim_{n to infty} y_n 0 ]

Step 4: Resolving the Original Limit

Recall that ( y_n ln left( n^{sqrt{frac{1}{n}}} right) ), and we have found that ( lim_{n to infty} y_n 0 ). To find the limit of the original sequence ( n^{sqrt{frac{1}{n}}} ), we exponentiate ( y_n ) with base ( e ) (the natural exponential function): [ lim_{n to infty} n^{sqrt{frac{1}{n}}} exp left( lim_{n to infty} y_n right) exp(0) 1 ]

Conclusion

In conclusion, we have rigorously proven that the sequence ( n^{sqrt{frac{1}{n}}} ) converges to 1 as ( n ) approaches infinity. The steps involved included taking the natural logarithm of the sequence, simplifying the expression, applying L'H?pital's Rule to resolve the indeterminate form, and finally exponentiating the result.

This detailed explanation not only provides a clear understanding of the process but also showcases the powerful tools of real analysis and calculus in addressing complex mathematical problems. Understanding such proofs is fundamental for advanced mathematical studies and applications.