Question

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} b^{-n}=0, \text { for } b > 1$$

Short answer

Question: Prove that if \(b > 1\), then \(\lim_{n \rightarrow \infty} b^{-n} = 0\). Answer: By using the formal definition of the limit and transforming the inequality \(b^{-n} < \epsilon\) to find a suitable natural number \(N\), we can prove that \(\lim_{n \rightarrow \infty} b^{-n} = 0\) for all \(b > 1\).

Step-by-step solution

Rewrite inequality

To begin, let's rewrite the inequality \(|b^{-n} - 0| < \epsilon\). Since \(b > 1\), we have \(b^{-n} > 0\), so the absolute value is not necessary: $$b^{-n} < \epsilon$$ Next, take the natural logarithm of both sides, which gives us: $$-n \ln(b) < \ln(\epsilon)$$ Since \(\ln(b) > 0\), we can divide by it: $$-n < \frac{\ln(\epsilon)}{\ln(b)}$$ Finally, we can multiply both sides by \(-1\) and arrive at the inequality: $$n>-\frac{\ln(\epsilon)}{\ln(b)}$$

Find N

Now we need to find a natural number \(N\) that satisfies the inequality \(n > -\frac{\ln(\epsilon)}{\ln(b)}\) for all \(n > N\). Define \(N = \lfloor -\frac{\ln(\epsilon)}{\ln(b)} \rfloor + 1\), where \(\lfloor x \rfloor\) denotes the largest integer less than or equal to \(x\). Since \(N\) is greater than \(-\frac{\ln(\epsilon)}{\ln(b)}\) and \(N\) is an integer, it's true that: $$n > -\frac{\ln(\epsilon)}{\ln(b)}, \forall n > N$$

Prove the limit

We want to prove that $$\lim_{n \rightarrow \infty} b^{-n} = 0$$ According to the formal definition, we need to find a natural number \(N\) that depends on \(\epsilon > 0\), such that the inequality \(b^{-n} < \epsilon\) holds for all \(n > N\). From Step 2, we found that for \(n > \lfloor -\frac{\ln(\epsilon)}{\ln(b)} \rfloor + 1 = N\), the inequality \(b^{-n} < \epsilon\) holds, so the formal definition of a limit is satisfied for every \(\epsilon > 0\). Therefore, we can conclude that: $$\lim_{n \rightarrow \infty} b^{-n} = 0 \text{, for } b > 1$$

Key concepts

Formal Definition of Limit

The formal definition of the limit of a sequence provides a rigorous way to say that a sequence gets arbitrarily close to a number, as the index goes off to infinity. For a sequence \(a_n\) to have a limit \(L\), we use the statement: "\(a_n\) approaches \(L\) as \(n\) approaches infinity". In formal terms, \(\lim_{n \to \infty} a_n = L\) if for every positive number \(\epsilon\), however small, there exists a natural number \(N\) such that for all \(n > N\), \(|a_n - L| < \epsilon\).*
This definition can be broken down into parts to better understand it:*

• The sequence \(a_n\) is assumed to come arbitrarily close to \(L\).
• \(|a_n - L|\) represents the distance between the term \(a_n\) and the limit \(L\).
• \(\epsilon\) can be thought of as a tolerance level. No matter how small this value is, we should be able to find a time \(N\) after which all terms \(a_n\) fall within this tolerance when \(n > N\).

This allows us to formally prove whether sequences like \(b^{-n}\) converge to a particular value when \(n\) becomes very large.

Natural Logarithm

The natural logarithm is a key mathematical function often denoted as \(\ln(x)\). It is the inverse function of the exponential function when the base of the logarithm is \(e\), where \(e\) is approximately equal to 2.71828. The function \(\ln(x)\) has several important properties:

• \(\ln(1) = 0\) because \(e^0 = 1\).
• The natural logarithm is only defined for positive numbers, meaning it does not work for zero or negative numbers.
• The derivative of \(\ln(x)\) is \(\frac{1}{x}\), which is useful in calculus and when dealing with growth problems.
• It can transform multiplicative processes into additive ones, which simplifies many algebraic operations.

In the context of limits, the natural logarithm helps by simplifying expressions during inequality manipulations. In the exercise, \(\ln(b)\) is particularly important because multiplying by it allows us to isolate \(n\) when showing that for \(b > 1\), \(b^{-n}\) approaches zero.

Inequality Manipulation

Inequality manipulation involves using algebraic rules to transform inequalities into more useful forms. This process is crucial when proving limits rigorously. Here's how it works:

• **Start with the Target Expression**: The inequality \(b^{-n} < \epsilon\) serves as our starting point.
• **Apply Algebraic Operations**: Depending on the context, operations such as logarithms might be applied. In this case, taking the natural logarithm of both sides gives us a manageable linear form:

\[ -n \ln(b) < \ln(\epsilon) \]

• **Rearranging and Solving**: Further manipulation involves dividing by constants (like \(\ln(b)\)) and multiplying or dividing by \(-1\). Considerations include being cautious about reversing inequality signs when multiplying or dividing by negatives.
• **Conclusion**: After manipulation, you'll have an expression, often involving \(n\), that supports finding a suitable \(N\) for proving limits. Here, such manipulation helps to show that \(n > -\frac{\ln(\epsilon)}{\ln(b)}\) must be satisfied.

This guides right to the solution, demonstrating how each step is essential in limiting the sequence.

Sequence Convergence

Sequence convergence refers to the tendency of a sequence to approach a specific value, called the limit, as the sequence progresses. A sequence \(a_n\) is said to converge to a limit \(L\) if the terms of the sequence get closer and closer to \(L\) as \(n\) increases.

• **Behavior of Terms**: In practical terms, the terms \(a_n\) of the sequence will "settle" near \(L\) and for any small positive number \(\epsilon\), you can find a point \(N\) after which all terms are within \(\epsilon\) distance of \(L\).
• **Convergence Example**: In our exercise, sequences of the form \(b^{-n}\) are shown to converge to 0 when \(b > 1\). This is done by demonstrating \(b^{-n}\) becomes negligibly small as \(n\) becomes very large.
• **Importance of \(N\)**: Finding \(N\) is key because it's the threshold beyond which the sequence's behavior is under control. For \(n > N\), a sequence that converges will consistently behave as expected.

Understanding sequence convergence means recognizing that no matter how the sequence starts or fluctuates initially, eventually, its terms will reliably approach the limit when properly proven.