Problem 55 - A Course in Calculus and Real Analysis

Question

For \(k \in \mathbb{N}\), let \(a_{k} \in \mathbb{R}\) with \(a_{k}>0 .\) Show that $$ \liminf _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} \leq \liminf _{k \rightarrow \infty} a_{k}^{1 / k} \quad \text { and } \quad \limsup _{k \rightarrow \infty} a_{k}^{1 / k} \leq \limsup _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} $$

Step-by-Step Solution

Answer

We showed that the following inequalities hold: $$ \liminf _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} \leq \liminf _{k \rightarrow \infty} a_{k}^{1 / k} \quad \text { and } \quad \limsup _{k \rightarrow \infty} a_{k}^{1 / k} \leq \limsup _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} $$ by defining the sequences \(b_k = \log(a_k)\) and \(c_k = \frac{a_{k+1}}{a_k}\), working with limit inferior and limit superior, using logarithms to simplify expressions, deriving relationships among the sequences, and combining inequalities to obtain the result.

Step 1: Understanding limit inferior and limit superior

Limit inferior and limit superior are ways to describe the limiting behavior of a sequence, even when it oscillates or doesn't have a unique limit. The limit inferior is the largest number that is the limit of infinitely many sub-sequences, while the limit superior is the smallest number that is the limit of infinitely many sub-sequences. In simple terms, the limit inferior represents the lowest possible limit, while the limit superior represents the highest possible limit if the sequence were infinitely "truncated" (by considering only subsequences).

Step 2: Use logarithms to simplify expressions

We can use logarithms to simplify the expressions in the limits. Let's define the sequence \(b_k = \log(a_k)\) and \(c_k = \frac{a_{k+1}}{a_k}\). Then the inequality we want to show becomes: $$ \liminf _{k \rightarrow \infty} \log(c_{k}) \leq \liminf _{k \rightarrow \infty} \frac{b_{k}}{k} \quad \text { and } \quad \limsup _{k \rightarrow \infty} \frac{b_{k}}{k} \leq \limsup _{k \rightarrow \infty} \log(c_{k}) $$

Step 3: Derive relationships among the sequences

We know that \(c_k = \frac{a_{k+1}}{a_k}\). Taking logarithms, we get: \(\log(c_k) = \log(a_{k+1}) - \log(a_k)\). Now we can rewrite the inequality we want to show as: $$ \liminf _{k \rightarrow \infty} (b_{k+1} - b_k) \leq \liminf _{k \rightarrow \infty} \frac{b_{k}}{k} \quad \text { and } \quad \limsup _{k \rightarrow \infty} \frac{b_{k}}{k} \leq \limsup _{k \rightarrow \infty} (b_{k+1} - b_k) $$

Step 4: Working with limit inferior

Let's first analyze the limit inferior inequality: $$\liminf_{k\rightarrow\infty}(b_{k+1}-b_k) \leq \liminf_{k\rightarrow\infty}\frac{b_k}{k}$$ For any given \(k\), we can sum the left side of the inequality from \(k=1\) to \(k=n\): $$\sum_{k=1}^n (b_{k+1} - b_k) = b_{n+1} - b_1$$ Now, divide by \(n\) and take the limit as \(n\) approaches infinity: $$\liminf_{n\rightarrow\infty} \frac{b_{n+1}}{n} - \liminf_{k\rightarrow\infty} \frac{b_1}{k} \leq \liminf_{k\rightarrow\infty}\frac{b_k}{k}$$ Since \(b_1\) is a constant, \(\liminf_{k\rightarrow\infty} \frac{b_1}{k} = 0\), and we have: $$\liminf_{n\rightarrow\infty} \frac{b_{n+1}}{n} \leq \liminf_{k\rightarrow\infty}\frac{b_k}{k}$$

Step 5: Working with limit superior

Now, let's analyze the limit superior inequality: $$\limsup_{k\rightarrow\infty}\frac{b_k}{k} \leq \limsup_{k\rightarrow\infty}(b_{k+1}-b_k)$$ We can perform the same summation as in Step 4 and obtain the following inequality: $$\limsup_{k\rightarrow\infty}\frac{b_k}{k} \leq \limsup_{n\rightarrow\infty} \frac{b_{n+1}}{n}$$

Step 6: Combine the inequalities to obtain the result

Now, let's combine the inequalities we derived from Steps 4 and 5: $$\liminf_{n\rightarrow\infty} \frac{b_{n+1}}{n} \leq \liminf_{k\rightarrow\infty}\frac{b_k}{k} \leq \limsup_{k\rightarrow\infty}\frac{b_k}{k} \leq \limsup_{n\rightarrow\infty} \frac{b_{n+1}}{n}$$ Finally, we can re-express the limits in terms of the original sequences \(a_k\) and \(c_k\), obtaining the desired result: $$ \liminf _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} \leq \liminf _{k \rightarrow \infty} a_{k}^{1 / k} \quad \text { and } \quad \limsup _{k \rightarrow \infty} a_{k}^{1 / k} \leq \limsup _{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} $$

Key Concepts

Sequence Convergence

When we talk about sequence convergence, we're referring to the idea that as a sequence progresses—think of it like a list of numbers that continues indefinitely—its terms get closer and closer to a specific value. This specific value is called the 'limit'. Think of it as you're playing a game of 'hot and cold', where 'hot' is picking numbers that are close to the limit, and 'cold' is picking numbers that are far away. The goal is to get 'hotter' indefinitely, meaning the numbers you pick become indistinguishable from the limit.
A sequence 'converges' if it has a limit. It's akin to saying, no matter how far down this list you go, the numbers keep getting closer to a particular value. If they do, bingo! That's convergence. When we put uncertainties aside and talk about limits in their most comfortable setting, it's when sequences behave nicely—with convergence, mathematical work becomes a streamlined process.
Monitoring sequence convergence is like watching a flower bloom; with the right conditions, it unfolds predictably and beautifully. In mathematics, we like sequences that converge because they're predictable and allow us to build upon them, creating more complex and exciting mathematical structures.

Inequalities in Calculus

In the world of calculus, inequalities help us understand the boundaries within which our mathematical expressions operate. Think of inequalities like the sidelines on a soccer field, establishing where the action can take place. In the context of calculus, rather than dealing with finite, sharp numbers, we often discuss ranges and intervals—picture this as playing soccer with time and space stretching and expanding.
Inequalities ensure that we can confidently talk about limits, as in the case with our limit inferior and limit superior discussion. They enable us to say, 'Even if we don't know the exact limit, we can say with certainty it won't go out of this range.' That's a profound and crucial concept in analysis, making sure mathematicians stay within the 'playing field' of logical reasoning while grappling with the infinite and the infinitesimal.
By detecting these boundaries, we can safely explore the vast ocean of calculus without fear of getting 'lost at sea'. They provide a compass that points mathematicians towards the correct shores of understanding, no matter how turbulent the mathematical seas may become.

Logarithms in Limits

Now, let's consider logarithms in limits. Logarithms are the inverse operation of exponentiation, just as subtraction is the inverse of addition. In the context of limits, logarithms can be a wily trick up our sleeve—they have a special talent for taming wild expressions involving exponents into pet-like obedience.
By introducing logarithms, daunting towering expressions shrink down to something more manageable, allowing us to better understand the behavior of the sequence as it stretches into infinity. Like a mathematical 'Honey, I Shrunk the Kids', logarithms reduce the size of the problem so that you can walk around it, inspect it, and solve it with fewer headaches.
Employing logarithms doesn't alter the core nature of the limit itself; it's like translating a foreign phrase into your native tongue. The meaning stays the same; it's just easier to understand. This is exactly what's been done in the exercise provided: using logarithms to transform the problem into a more digestible format, which then allows us to analyze the sequence's behavior as it approaches infinity.