Problem 15 - Basic Analysis: Introduction to Real Analysis

Question

Prove \(\lim _{n \rightarrow \infty}\left(n^{2}+1\right)^{1 / n}=1\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Limit Expression

We are asked to prove that \(\lim _{n \rightarrow \infty}\left(n^{2}+1\right)^{1 / n}=1\). Essentially, we want to find the behavior of the sequence \((n^2 + 1)^{1/n}\) as \(n\) approaches infinity.

Step 2: Analyze Using Logarithms

To simplify the expression, take the natural logarithm. Let \(a_n = (n^2 + 1)^{1/n}\), then \(\ln(a_n) = \frac{1}{n} \ln(n^2 + 1)\). Our goal is to show that \(\ln(a_n)\) approaches 0 as \(n\) goes to infinity.

Step 3: Evaluate the Logarithm

Simplify \(\ln(n^2 + 1)\) by using expansion. We have: \(\ln(n^2 + 1) = \ln(n^2 (1 + \frac{1}{n^2})) = 2\ln(n) + \ln(1 + \frac{1}{n^2})\).For large \(n\), we use the fact that \(\ln(1 + x) \approx x\) for small \(x\). Thus, \(\ln(1 + \frac{1}{n^2}) \approx \frac{1}{n^2}\).

Step 4: Substitute and Simplify

Substitute back into the expression for \(\ln(a_n)\): \(\ln(a_n) = \frac{1}{n} (2 \ln(n) + \frac{1}{n^2})\).The term \(\frac{1}{n^3}\) approaches 0 faster than \(\frac{\ln(n)}{n}\), so focus on \(\frac{2 \ln(n)}{n}\).

Step 5: Determine the Limit of Each Part

We need to evaluate \(\lim_{n\to\infty} \frac{2 \ln(n)}{n}\). Using L'Hopital's Rule, since both numerator and denominator approach infinity, differentiating yields \(\frac{2/n}{1} = \frac{2}{n}\). As \(n \to \infty\), \(\frac{2}{n} \to 0\).

Step 6: Conclude the Proof

Therefore, \(\ln(a_n) \to 0\) and since the exponential function is continuous, it implies \(a_n = e^{\ln(a_n)} \to e^0 = 1\). We have thus proven that \(\lim_{n \to \infty} (n^2 + 1)^{1/n} = 1\).

Key Concepts

Natural logarithm

Natural logarithms are an essential tool when analyzing sequences, especially when dealing with expressions containing exponents. They help break down complex problems into simpler, more manageable parts.

A natural logarithm is the logarithm to the base of the mathematical constant \(e\), approximately equal to 2.71828. We denote it as \(\ln(x)\). When you see a term like \((n^2 + 1)^{1/n}\), taking the natural logarithm turns multiplication and division into addition and subtraction. This is why \(\ln((n^2 + 1)^{1/n})\) becomes \(\frac{1}{n} \ln(n^2 + 1)\).

By transforming the expression using natural logarithms, one can more easily evaluate limits and understand the growth behavior of sequences.

L'Hôpital's rule

When faced with indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's rule provides a method for finding limits. It involves differentiating the numerator and the denominator separately and then taking the limit again.

In the problem, as \(n\) approaches infinity, the term \(\frac{2\ln(n)}{n}\) evaluated to \(\frac{\infty}{\infty}\). L'Hôpital's rule lets us differentiate: differentiate \(2\ln(n)\) to get \(\frac{2}{n}\) and differentiate \(n\) to get \(1\). Thus, the expression becomes \(\frac{2/n}{1}\).

Finally, as \(n\) moves towards infinity, \(\frac{2}{n}\) approaches 0. This method provides a simple route to evaluate complicated limits.

Exponential function

The exponential function, often denoted as \(e^x\), is the inverse of the natural logarithm function. It is a continuous and smooth function that, when applied to the result of a logarithmic evaluation of a sequence, can help determine the sequence's limit behavior.

In the context of this problem, after proving \(\ln(a_n)\) tends to 0, we utilize the exponential function to revert from the logarithmic space back to the original expression. Since \(e^{\ln(a_n)} = a_n\), and \(\ln(a_n) \to 0\), it implies \(a_n \to e^0 = 1\).

This step is crucial because it verifies through the continuity of the exponential function that the convergence we have shown in the logarithmic form translates to the convergence in the original sequence.

Sequence convergence

Understanding sequence convergence is central to analyzing limits. A sequence converges if its terms approach a specific value as the sequence progresses to infinity. In simple terms, a convergent sequence settles into a particular pattern or value over time.

For the sequence \((n^2 + 1)^{1/n}\), convergence means as \(n\) increases, the sequence's value approaches 1. In this exercise, we demonstrated convergence by manipulating the sequence using transformations like the natural logarithm and using mathematical techniques like L'Hôpital's rule.

Ultimately, proving that the sequence converges to 1 entails confirming that all transformations and evaluations, including exponential and logarithmic adjustments, indicate a consistent approach to that limit. This is a powerful demonstration of how various mathematical concepts interlink to determine the behavior of sequences.