Problem 15 - Advanced Calculus

Question

Let \(\left\\{a_{n}\right\\}\) be a sequence of real numbers. Suppose that for each positive number \(c\) there is an index \(N\) such that $$a_{n}>c$$ for all indices \(n \geq N\) When this is so, the sequence \(\left\\{a_{n}\right\\}\) is said to converge to infinity, and we write $$\lim _{n \rightarrow \infty} a_{n}=\infty$$ Prove the following: a. \(\lim _{n \rightarrow \infty}\left[n^{3}-4 n^{2}-100 n\right]=\infty\) $$\text { b. } \lim _{n \rightarrow \infty}\left[\sqrt{n}-\frac{1}{n^{2}}+4\right]=\infty$$

Step-by-Step Solution

Answer

Both sequences grow without bound as \(n\) goes to infinity. For part (a), \(n^3\) eventually dominates, and for part (b), \(\sqrt{n}\) dominates, ensuring each sequence is greater than any positive \(c\) for \(n \geq N\).

Step 1 - Understand the Definition of Convergence to Infinity

To show that the sequence converges to infinity, we need to demonstrate that for every positive real number c, there is an index (called N) beyond which every term in the sequence is greater than c. This means after some point in the sequence, all the terms will be larger than any given positive number.

Step 2 - Consider the Cubic Term Dominance

For part (a), we notice that as n becomes very large, the term \(n^3\) will dominate the behavior of the sequence because it grows faster than \(n^2\) and \(n\). This means that eventually, the \(n^3\) term will make the sequence very large, no matter the values of the other terms.

Step 3 - Show the Sequence Part (a) is Greater than c

We show that for any positive c, we can find N such that for all \(n \geq N\), \(a_n = n^3 - 4n^2 - 100n > c\). As n gets larger, the \(n^3\) term will be much larger than the combined \(4n^2 + 100n\) terms, hence making \(a_n\) larger than c.

Step 4 - Consider the Square Root Term Dominance

For part (b), we notice that as n becomes large, the \(\sqrt{n}\) term grows without bounds, whereas \(\frac{1}{n^2}\) shrinks towards zero. Therefore, the growth of \(\/sqrt{n}\) term will make the whole sequence grow indefinitely.

Step 5 - Show the Sequence Part (b) is Greater than c

We show that for any positive c, there exists an N such that for all \(n \geq N\), \(b_n = \sqrt{n} - \frac{1}{n^2} + 4 > c\). The term \(\sqrt{n}\) will eventually outweigh the \(-\frac{1}{n^2}\) term and the sequence will grow larger than any positive number c because \(\/sqrt{n}\) increases without bound as n increases.

Key Concepts

Limits of Sequences

The concept of a sequence approaching a limit is fundamental in calculus and analysis. A limit of a sequence is a value that the elements of the sequence get increasingly close to as the sequence progresses. When we say a sequence \(\left\{a_n\right\}\) converges to infinity, we're describing a scenario where, after a certain point, all the terms in the sequence grow beyond any given positive number. In our exercise, this is formalized as:\[\lim _{n \rightarrow \infty} a_{n}=\infty\]

In order to prove such a statement, one must follow the definition closely, showing that for any chosen positive real number \(c\), there is a point in the sequence, labeled \(N\), beyond which all terms of the sequence exceed \(c\). The goal is to specify that no matter how big \(c\) is chosen, the terms of the sequence will eventually surpass it, affirming that the sequence grows without bounds.

Dominance in Polynomial Terms

In examining the behavior of polynomials, it becomes clear that terms with higher exponents will eventually outpace those with lower exponents as the variable grows larger. This is referred to as 'dominance'. In a polynomial, the term with the highest degree will dominate the growth of the polynomial for large values of the variable. In our exercise's part (a), \(n^3\) dominates over \(n^2\) and \(n\), ensuring that the polynomial \(n^3 - 4n^2 - 100n\) will grow indefinitely as \(n\) increases.

Understanding this concept aids in evaluating limits that involve polynomial expressions. By focusing on the dominant term, we can simplify the problem and make a reasonable assertion about the behavior of the sequence as \(n\) approaches infinity. For students, grasping dominance is key to predicting the long-term behavior of polynomials without resorting to complex calculations.

Behavior of Sequences

Sequences can exhibit various behaviors as their terms progress towards infinity. For instance, some sequences converge to a specific number, oscillate between values, or, as highlighted in part (b) of our exercise, grow without bound. The behavior of the sequence \(\sqrt{n} - \frac{1}{n^2} + 4\) can be understood by examining its individual components. The term \(\sqrt{n}\) increases without limit as \(n\) grows, while \(\frac{1}{n^2}\) diminishes towards zero. The constant term \(4\) has no effect on the sequence's progression towards infinity.

This evaluation allows us to determine that the sequence will eventually exceed any positive value \(c\), and thus, can be said to converge to infinity. This understanding of behavior provides a foundation for exploring more complex sequences and series, enhancing students' analytical capabilities in mathematics.