Question

Prove that convergence of \(\left\\{s_{n}\right\\}\) implies convergence of \(\left\\{\left|s_{n}\right|\right\\} .\) Is the converse true? Calculate \(\lim \left(\sqrt{n^{2}+n}-n\right)\).

Short answer

Is the converse true? Calculate the limit \(\lim\left(\sqrt{n^{2}+n}-n\right)\) as \(n\) approaches infinity. Answer: If a sequence \(s_n\) converges, the sequence of its absolute values \(|s_n|\) also converges. However, the converse is not true. The limit \(\lim\left(\sqrt{n^{2}+n}-n\right)\) as \(n\) approaches infinity is \(\frac{1}{2}\).

Step-by-step solution

Prove that convergence of \(s_n\) implies convergence of \(|s_n|\)

Let us assume that the sequence \(s_n\) converges to a limit \(L\). That is, for any \(\varepsilon > 0\), there exists an integer \(N\) such that for all \(n > N\), we have \(|s_n - L| < \varepsilon\). Now consider the sequence of absolute values, \(|s_n|\). Our goal is to prove that this sequence also converges to \(|L|\). For any \(\varepsilon > 0\), we need to find an integer \(M\) such that for all \(n > M\), we have \(||s_n| - |L|| < \varepsilon\). Notice that by the triangle inequality, we have: $$||s_n| - |L|| \leq |s_n - L|$$ Since \(|s_n - L| < \varepsilon\) for all \(n > N\), it is clear that \(||s_n| - |L|| < \varepsilon\) for all \(n > N\). Thus, the sequence of absolute values \(|s_n|\) converges to \(|L|\).

Determine if the converse is true

Now let's consider the converse: If the sequence of absolute values \(|s_n|\) converges, does that imply that the original sequence \(s_n\) also converges? To answer this question, consider the following counterexample: Let \(s_n = (-1)^n\). The sequence of absolute values is \(|s_n| = 1\) for all \(n\). This is a constant sequence, which clearly converges to \(1\). However, the original sequence \(s_n = (-1)^n\) does not converge, as it oscillates between \(-1\) and \(1\). Therefore, the converse is not true.

Calculate \(\lim\left(\sqrt{n^{2}+n}-n\right)\)

To calculate \(\lim\left(\sqrt{n^{2}+n}-n\right)\) as \(n\) approaches infinity, we can try to factor the expression inside the square root: $$\lim_{n\to\infty} \left(\sqrt{n^2 + n} - n\right) = \lim_{n\to\infty} \left(\sqrt{n(n+1)} - n\right)$$ Now, let's divide and multiply the expression inside the square root by \(\left(\sqrt{n(n+1)} + n\right)\): $$\lim_{n\to\infty} \left(\sqrt{n^2 + n} - n\right) = \lim_{n\to\infty} \frac{n(n+1)-n^2}{\sqrt{n(n+1)} + n}$$ Simplifying the numerator, we get: $$\lim_{n\to\infty} \frac{n}{\sqrt{n(n+1)} + n}$$ Now divide both numerator and denominator by \(n\): $$\lim_{n\to\infty} \frac{1}{\sqrt{1+\frac{1}{n}} + 1}$$ As \(n\) approaches infinity, the fraction \(\frac{1}{n}\) approaches \(0\). Thus, we have: $$\lim_{n\to\infty} \frac{1}{\sqrt{1+\frac{1}{n}} + 1} = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{2}$$ So, \(\lim\left(\sqrt{n^{2}+n}-n\right) = \frac{1}{2}\).

Key concepts

Convergence of Sequences

In mathematics, a sequence is said to converge if it approaches a specific value, known as its limit, as the sequence progresses to infinity. Let's break it down: if you have a sequence \(s_n\) that converges to a limit \(L\), it means that the terms of the sequence get closer and closer to \(L\) as \(n\) increases.

• For any given small distance \(\varepsilon > 0\), there exists a point \(N\) beyond which all terms \(s_n\) satisfy \(|s_n - L| < \varepsilon\).

This is a key property in calculus and analysis since it allows us to precisely define how sequences behave as they extend into infinity. By proving this convergence, one can solve problems related to function behavior, continuity, and more.

Absolute Value of Sequences

The absolute value of a sequence, written as \(|s_n|\), modifies the sequence such that all its terms are non-negative. If the sequence \(s_n\) itself converges, its absolute value \(|s_n|\) will also converge. The convergence of the original sequence implies that its values stabilize - they don't vary much irrespective of whether they're positive or negative.

The Role of the Triangle Inequality

By using the triangle inequality: \[||s_n| - |L|| \leq |s_n - L|\]We can conclude that the sequence \(|s_n|\) converges to \(|L|\). This property holds due to the nature of the absolute value, which flattens out oscillations and enables the comparison of sequence terms to their limit. Thus, the absolute value preserves the convergence property, demonstrating that the stabilization effect is not lost when the absolute value is applied.

Limits at Infinity

Calculating limits at infinity involves evaluating what the terms of a sequence or the value of a function approach as the variable grows indefinitely large. One commonly uses algebraic manipulation to simplify expressions, as seen in the following example: When tasked with evaluating \(\lim(\sqrt{n^2 + n} - n)\), we employed factorization and division techniques to isolate dominant terms.

• By rewriting expressions and dividing both numerator and denominator by \(n\), simplification becomes possible.

In this process:1. Clear up complex terms so their limiting behavior is made evident.2. Simplify the expression to identify which parts diminish as \(n\) grows.This approach is elemental in calculus, providing the groundwork needed to underpin more complex limits and advanced calculus concepts.

Counterexamples in Convergence

In convergence, not all intuitive assumptions hold. The importance of counterexamples lies in validating mathematical concepts and proving the necessity of conditions.Consider the sequence \(s_n = (-1)^n\), representing an alternating set of terms \(-1, 1, -1, 1,\dots\) whose absolute value \(|s_n|=1\) converges to \(1\). However, the sequence \(s_n\) itself does not converge.

• The terms of \(s_n\) keep oscillating; therefore, no single limit exists.

Counterexamples demonstrate that the apparent convergence of absolute values does not necessarily imply the original sequence converges too. Hence, they serve as crucial tools to refute mistaken mathematical assumptions, guiding learners to more accurate conclusions.