Problem 43 - 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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{(-1)^{n}}{n+1} $$

Step-by-Step Solution

Answer

The first five terms are 1, -1/2, 1/3, -1/4, 1/5 and the limit is 0.

Step 1: Calculate the First Term

To find the first term of the sequence, substitute \(n = 0\) into the formula. \[a_0 = \frac{(-1)^0}{0+1} = \frac{1}{1} = 1.\]

Step 2: Calculate the Second Term

Substitute \(n = 1\) into the formula to find the second term. \[a_1 = \frac{(-1)^1}{1+1} = \frac{-1}{2}.\]

Step 3: Calculate the Third Term

For the third term, substitute \(n = 2\). \[a_2 = \frac{(-1)^2}{2+1} = \frac{1}{3}.\]

Step 4: Calculate the Fourth Term

To find the fourth term, substitute \(n = 3\). \[a_3 = \frac{(-1)^3}{3+1} = \frac{-1}{4}.\]

Step 5: Calculate the Fifth Term

Finally, for the fifth term, substitute \(n = 4\). \[a_4 = \frac{(-1)^4}{4+1} = \frac{1}{5}.\]

Step 6: Determine the Limit

To find the limit of the sequence as \(n\) approaches infinity, consider the expression \(\frac{(-1)^n}{n+1}\). As \(n\) increases, the denominator \(n+1\) becomes very large, so the fraction \(\frac{1}{n+1}\) approaches 0. Thus, \(\lim_{n \to \infty} a_n = 0.\)

Key Concepts

Sequence Terms Calculation

Calculating sequence terms involves using a given formula to find each term by substituting values for the sequence index, often labeled as \( n \). In the exercise provided, the sequence is defined by the formula \( a_n = \frac{(-1)^n}{n+1} \). To find each term:

• Start by substituting \( n = 0 \) to find the first term: \( a_0 = \frac{1}{1} = 1 \).


• Substitute \( n = 1 \) for the second term: \( a_1 = \frac{-1}{2} \).


• Continue this process for \( n = 2,\;3, \) and \( 4 \) to find: \( a_2 = \frac{1}{3}, \; a_3 = \frac{-1}{4}, \) and \( a_4 = \frac{1}{5} \).

This iterative method provides a systematic approach to determine each sequence term. Calculating these terms visibly shows how the alternating nature and decreasing magnitude play out in the sequence.

Infinite Limits

Determining the limit of a sequence as \( n \) approaches infinity is crucial in understanding the sequence's behavior in the long run. For the given sequence \( a_n = \frac{(-1)^n}{n+1} \), observe how the fraction behaves as \( n \) increases indefinitely.

• The term \( (-1)^n \) makes the sequence alternate between positive and negative.


• The denominator \( n+1 \) grows ever larger, causing the fraction's absolute value \( \frac{1}{n+1} \) to get closer to 0.

Despite the sequence's oscillations due to the \((-1)^n\) factor, the primary influence on the sequence's limit is the denominator. As \( n \) increases without bound, \( \frac{1}{n+1} \) approaches 0; thus, the overall limit is \( \lim_{n \to \infty} a_n = 0 \). Understanding this infinite limit helps us conclude the sequence trends toward 0 despite its alternation in signs.

Alternating Sequences

Sequences that switch between positive and negative terms are known as alternating sequences. These sequences have specific characteristics that influence their pattern and overall limit behavior.

• The term \((-1)^n\) is essential in creating the alternating pattern. For even \( n \), \((-1)^n = 1 \), and for odd \( n \), \((-1)^n = -1 \).


• In the sequence \( a_n = \frac{(-1)^n}{n+1} \), this factor means the sequence flips signs with each consecutive term.

Alternating sequences are interesting because their oscillations between positive and negative values can converge to a specific limit, as seen in this case where the sequence approaches 0. These sequences provide good practice in understanding both their pattern through direct calculation and their longer-term behavior through limit evaluation.