Problem 42 - 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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{1}{\sqrt{n+1}} $$

Step-by-Step Solution

Answer

The first five terms are 1, \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{2}\), \(\frac{1}{\sqrt{5}}\), and the limit is 0.

Step 1: Find the First Term

To find the first term of the sequence, substitute \(n = 0\) into the formula \(a_{n} = \frac{1}{\sqrt{n+1}}\). This gives:\[a_0 = \frac{1}{\sqrt{0+1}} = \frac{1}{\sqrt{1}} = 1.\]

Step 2: Find the Second Term

To find the second term, substitute \(n = 1\) into the formula. This gives:\[a_1 = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}.\]

Step 3: Find the Third Term

To find the third term, substitute \(n = 2\) into the formula. This gives:\[a_2 = \frac{1}{\sqrt{2+1}} = \frac{1}{\sqrt{3}}.\]

Step 4: Find the Fourth Term

To find the fourth term, substitute \(n = 3\) into the formula. This gives:\[a_3 = \frac{1}{\sqrt{3+1}} = \frac{1}{\sqrt{4}} = \frac{1}{2}.\]

Step 5: Find the Fifth Term

To find the fifth term, substitute \(n = 4\) into the formula. This gives:\[a_4 = \frac{1}{\sqrt{4+1}} = \frac{1}{\sqrt{5}}.\]

Step 6: Find the Limit as n Approaches Infinity

To find \(\lim _{n \rightarrow \infty} a_{n}\), consider the behavior of \(a_n\):\[a_n = \frac{1}{\sqrt{n+1}}.\]As \(n\) becomes very large, \(\sqrt{n+1}\) becomes very large. Hence, \(\frac{1}{\sqrt{n+1}}\) approaches 0.\[\lim _{n \rightarrow \infty} a_n = 0.\]

Key Concepts

Limit of a Sequence

In calculus, the limit of a sequence is a fundamental concept that helps us understand how a sequence behaves as the term number increases indefinitely. A sequence is an ordered list of numbers. The limit of a sequence as the index (often denoted \(n\)) approaches infinity, is a value that the sequence elements get closer to as \(n\) becomes very large.

Consider the sequence \(a_n = \frac{1}{\sqrt{n+1}}\). As \(n\) increases,

• \(\sqrt{n+1}\) also increases, causing the denominator to grow.
• Thus, \(\frac{1}{\sqrt{n+1}}\), our sequence term, gets smaller.

This shrinking behavior happens because the sequence approaches a value of zero.

Formally, it is expressed as \(\lim _{n \rightarrow \infty} a_n = 0\). Evaluating limits helps us predict long-term values for infinite sequences without calculating every single term.

Infinite Series

An infinite series is built upon the idea of sequences, but it goes one step further. It involves the summation of an infinite sequence of terms. Infinite series are pivotal in calculus, as they allow us to explore concepts like convergence and divergence, reflecting whether an infinite sum approaches a finite limit.

This idea can be tied back to the given sequence. While our primary focus was just on evaluating the limit of the sequence \(a_n = \frac{1}{\sqrt{n+1}}\), an infinite series would involve summing up all possible terms of this sequence:

\[ \sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}} \]

However, infinite series can be tricky. Not all series converge to a finite number. Some diverge, meaning their sum grows indefinitely. Understanding whether a series converges or diverges is crucial in advanced calculus topics.

Sequence Terms

Sequence terms are individual elements of a sequence. Each sequence term is typically generated by a specific formula or rule as seen in the exercise formula \(a_n = \frac{1}{\sqrt{n+1}}\). Each term corresponds to a specific \(n\) value.

Let's illustrate with the first five terms here:

• \(a_0 = 1\)
• \(a_1 = \frac{1}{\sqrt{2}}\)
• \(a_2 = \frac{1}{\sqrt{3}}\)
• \(a_3 = \frac{1}{2}\)
• \(a_4 = \frac{1}{\sqrt{5}}\)

Each term reflects the sequence's progression as \(n\) increases. Initially, for smaller values of \(n\), sequence terms can vary greatly in size.

But as calculated in the limit section, as \(n\) continues to increase, the impact of \(n\) becomes more pronounced, and terms approach a consistent behavior, leading to a limit.