Problem 51 - 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 \(\lim _{n \rightarrow \infty} a_{n^{*}}\) $$ a_{n}=\frac{(-1)^{n+1}}{n+1} $$

Step-by-Step Solution

Answer

The first five terms are -1, 0.5, -0.333, 0.25, -0.2, 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: \[ a_0 = \frac{(-1)^{0+1}}{0+1} = \frac{-1}{1} = -1 \]

Step 2: Calculate the second term

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

Step 3: Calculate the third term

Substitute \( n = 2 \) into the formula to find the third term: \[ a_2 = \frac{(-1)^{2+1}}{2+1} = \frac{-1}{3} \]

Step 4: Calculate the fourth term

Substitute \( n = 3 \) into the formula to find the fourth term: \[ a_3 = \frac{(-1)^{3+1}}{3+1} = \frac{1}{4} \]

Step 5: Calculate the fifth term

Substitute \( n = 4 \) into the formula to find the fifth term: \[ a_4 = \frac{(-1)^{4+1}}{4+1} = \frac{-1}{5} \]

Step 6: Identify the pattern in the sequence

The first five terms are: \(-1, \frac{1}{2}, \frac{-1}{3}, \frac{1}{4}, \frac{-1}{5}\). The sequence alternates signs and the magnitude of each term is the reciprocal of an increasing integer.

Step 7: Determine the limit of the sequence as \( n \rightarrow \infty \)

As \( n \) approaches infinity, the denominator \( n+1 \) becomes very large, making \( \frac{(-1)^{n+1}}{n+1} \) approach zero, because the terms \( \frac{1}{n+1} \) shrink towards zero.

Key Concepts

Understanding Sequence Terms

A sequence is essentially a list of numbers following a particular pattern. The numbers listed are referred to as the 'terms' of the sequence. Consider our sequence \(\{a_n\}\), where \(n\) represents the position of a term. The general formula given in the problem is \(a_n = \frac{(-1)^{n+1}}{n+1}\). This formula tells us how to find each term by plugging in different values of \(n\).

Here are the main points:

• The numerator \((-1)^{n+1}\) indicates that the signs of the terms alternate between positive and negative.
• The denominator \(n+1\) is the successive integer values starting from 1, which determines the division.'
  
  

By substituting different \(n\) values starting from 0, you can see this sequence's alternating signs and reciprocal pattern of integers.

The Limit of a Sequence

The limit of a sequence is the value that the sequence's terms approach as you proceed infinitely through the sequence. In simpler terms, if you keep calculating terms using the sequence's formula, you'll notice that they get closer and closer to a certain number. This number is called the 'limit.'

For the sequence \(a_n = \frac{(-1)^{n+1}}{n+1}\), as \(n\) becomes infinitely large, the denominator \(n+1\) increases significantly, making each term closer to zero. This is because as the denomintor goes up, the fraction as a whole gets smaller.

Thus, the limit of this sequence as \(n \to \infty\) is 0.

Understanding Convergence

Convergence refers to the behavior of a sequence in terms of its limit. A sequence is said to converge if its terms tend to approach a single finite limit as \(n \to \infty\). Essentially, it's like asking whether the series of numbers gets "sandwiched" or "pinned down" to a specific value when you keep going further down the line.

In our given sequence, \(a_n = \frac{(-1)^{n+1}}{n+1}\), the terms oscillate in sign but fade in magnitude, coming closer to zero. Because these terms settle at zero as \(n\) increases, the sequence converges, specifically to the number 0.

Remember these facts about convergence:

• A sequence that converges always approaches a real number as its limit.
• If terms get smaller and smaller, eventually approaching zero or any fixed number, they are exhibiting convergence.