Problem 52 - Calculus for Biology and Medicine

Question

Write the first five terms of the sequence \(\left\\{a_{n}\right\\}\), \(n=0,1,2,3, \ldots\), and find \(\lim _{n \rightarrow \infty} a_{n^{*}}\) $$ a_{n}=\frac{(-1)^{n}}{n^{3}+3} $$

Step-by-Step Solution

Answer

The first five terms are \(\frac{1}{3}, \frac{-1}{4}, \frac{1}{11}, \frac{-1}{30}, \frac{1}{67}\) and \(\lim_{n \to \infty} a_n = 0\).

Step 1: Calculate the First Term

To find the first term \(a_0\), substitute \(n=0\) into the formula \(a_n = \frac{(-1)^n}{n^3 + 3}\). Thus, \(a_0 = \frac{(-1)^0}{0^3 + 3} = \frac{1}{3}\).

Step 2: Calculate the Second Term

Find \(a_1\) by substituting \(n=1\), yielding \(a_1 = \frac{(-1)^1}{1^3 + 3} = \frac{-1}{4}\).

Step 3: Calculate the Third Term

Substitute \(n=2\) into the equation, which gives \(a_2 = \frac{(-1)^2}{2^3 + 3} = \frac{1}{11}\).

Step 4: Calculate the Fourth Term

Find \(a_3\) by substituting \(n=3\), resulting in \(a_3 = \frac{(-1)^3}{3^3 + 3} = \frac{-1}{30}\).

Step 5: Calculate the Fifth Term

Substitute \(n=4\) to get \(a_4 = \frac{(-1)^4}{4^3 + 3} = \frac{1}{67}\).

Step 6: Analyze the Limit as n Approaches Infinity

To find \(\lim_{n \to \infty} a_n\), observe the formula \(a_n = \frac{(-1)^n}{n^3 + 3}\). As \(n\) increases, \(n^3 + 3\) also increases, so \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(-1)^n}{n^3 + 3} = 0\) because the denominator grows much larger than the numerator.

Key Concepts

Sequence Terms

When dealing with sequences, it's important to understand that each element, or term, in the sequence follows a specific pattern or rule. For our given sequence \(a_n = \frac{(-1)^n}{n^3 + 3}\), each term is determined by plugging the value of \(n\) into this formula. In sequences, \(n\) typically starts at 0 and moves upwards in whole number increments (e.g., 0, 1, 2, 3, ...).

By substituting different values of \(n\) into this rule, you can generate each term:

• For \(n = 0\), the term is \(\frac{1}{3}\).
• For \(n = 1\), the term becomes \(\frac{-1}{4}\).
• For \(n = 2\), we get \(\frac{1}{11}\).
• At \(n = 3\), the term is \(\frac{-1}{30}\).
• And for \(n = 4\), it's \(\frac{1}{67}\).

It's important to carefully follow the pattern indicated by the formula, as it directs us how each term alternates and by which factor or fraction they are divided.

Convergence

Convergence refers to whether a sequence approaches a specific value, or limit, as the number of terms increases indefinitely. When a sequence converges, the values of the terms get closer to a fixed number as \(n\) approaches infinity.

In the case of \(a_n = \frac{(-1)^n}{n^3 + 3}\), as \(n\) gets larger, the denominator \(n^3 + 3\) increases significantly, making the fraction very small. Essentially, the effect of the numerator \((-1)^n\), which oscillates between -1 and +1, becomes negligible.

Thus, as \(n\) grows, \(a_n\) approaches 0. We express this mathematically as the limit: \(\lim_{n \to \infty} a_n = 0\). This means the sequence converges towards 0.

Alternating Series

An alternating series is a sequence where the sign of the terms alternates between positive and negative. This is evident in our sequence \(a_n = \frac{(-1)^n}{n^3 + 3}\), where the \((-1)^n\) factor ensures this alternation.

For odd values of \(n\) (like 1, 3, 5...), \((-1)^n\) becomes -1, resulting in negative terms. Conversely, for even \(n\) (such as 0, 2, 4...), the factor becomes +1, yielding positive terms. Hence, the sequence flips back and forth from positive to negative.

• This alternating nature can make the behavior of the series intriguing. It affects how quickly the terms approach their limit, contributing to the convergence process.
• An alternating series converges absolutely if the absolute values of the terms decrease steadily and approach zero.

In our sequence, despite the alternation, the convergence towards zero is not affected, primarily because as \(n\) increases, the denominator grows much faster than the numerator switches sign, driving the limit towards zero.