Question

In the following problems, find the limit of the given sequence as \(n \rightarrow \infty\). $$\frac{(-1)^{n} \sqrt{n+1}}{n}$$

Short answer

0

Step-by-step solution

Understand the Given Sequence

The given sequence is \( a_n = \frac{(-1)^{n} \sqrt{n+1}}{n} \). The goal is to find \( \lim_{n \to \infty} a_n \).

Simplify the Expression

Rewrite the sequence to separate the terms: \( \frac{(-1)^{n} \sqrt{n+1}}{n} = (-1)^{n} \cdot \frac{\sqrt{n+1}}{n} \).

Simplify \( \frac{\sqrt{n+1}}{n} \)

Divide both the numerator and the denominator inside the square root by \(n\): \( \frac{\sqrt{n+1}}{n} = \frac{\sqrt{n(1+\frac{1}{n})}}{n} = \frac{\sqrt{n} \cdot \sqrt{1+\frac{1}{n}}}{n} \).

Further Simplify

Notice that \( \sqrt{n} = n^{1/2} \), so \( \frac{\sqrt{n} \cdot \sqrt{1+\frac{1}{n}}}{n} = \frac{n^{1/2} \cdot \sqrt{1+\frac{1}{n}}}{n} = \frac{\sqrt{1+\frac{1}{n}}}{n^{1/2}} \).

Evaluate the Limit

As \( n \) approaches infinity, \( \frac{1}{n} \) approaches 0, so \( \sqrt{1+\frac{1}{n}} \) approaches \( \sqrt{1} = 1 \). The expression thus simplifies to \( \frac{1}{n^{1/2}} = \frac{1}{\sqrt{n}} \).

Find the Limit

The limit \( \lim_{n \to \infty} \frac{1}{\sqrt{n}} \) is 0 because \( \sqrt{n} \) grows without bound as \( n \) approaches infinity. Since the sequence is \( (-1)^{n} \) times a term that goes to 0, the whole sequence goes to 0 as \( n \) approaches infinity.

Key concepts

Infinite Limit

When we talk about finding the limit of a sequence as it approaches infinity, we are looking at the behavior of the sequence as the number of terms gets very large. For instance, in the sequence given in the exercise, we need to find what happens to the term \(\frac{(-1)^{n} \sqrt{n+1}}{n}\) as \(n\) gets larger and larger.

The essence of infinite limits lies in the fact that we want to see if the terms of the sequence get closer to a certain value, diverge to positive or negative infinity, or oscillate without settling down.

In the problem, to address the limit, we needed to simplify the sequence first. After simplification, our goal was to observe the term's behavior as \(n\) increases indefinitely. Through step-by-step analysis, it became clear that the given sequence tends towards 0 as \(n \to \infty\).

Sequence Simplification

Simplifying a sequence is often key to finding its limit. In the given exercise, we first separate the terms for easier manipulation. The sequence: \(\frac{(-1)^{n} \sqrt{n+1}}{n}\) was split into \( (-1)^{n} \times \frac{\sqrt{n+1}}{n}\).

To make further simplifications, we focused on simplifying \( \frac{\sqrt{n+1}}{n}\).

This was done by expressing both the numerator and the denominator inside the square root in terms of \(n \).

Notice that \(\frac{\sqrt{n+1}}{n}\ = \frac{\sqrt{n(1+\frac{1}{n})}}{n} \). By dividing and using the properties of square roots, we proceed step-by-step until reaching a form where we can easily identify the limit as \(n\) increases. This method simplifies complex-looking fractions and square roots, making it possible to determine the sequence behavior as variables grow larger.

Square Root Properties

Understanding square root properties is crucial in solving limits involving roots. In our example, knowing that \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \) helps in breaking down and simplifying expressions.

Let's consider the expression \(\frac{\sqrt{n+1}}{n}\) again. This can be simplified as \(\frac{\sqrt{n \times (1 + \frac{1}{n})}}{n} \). Applying the property that \(\frac{\sqrt{a}}{b} = \frac{a^{1/2}}{b} \) simplifies it further. Using \( \sqrt{n} = n^{1/2} \), the term transforms into \(\frac{n^{1/2} \cdot \sqrt{1 + \frac{1}{n}}}{n} \).

This further reduces to \(\frac{\sqrt{1 + \frac{1}{n}}}{n^{1/2}}\), which is easier to handle as \(n\) goes to infinity.

Knowing these properties allows one to reduce complex expressions step-by-step. It also shows how values that initially seem difficult to handle can simplify dramatically, leading to finding the limit smoothly.