Sequence Convergence: ∑(aₙ)ʳ Analysis

Hey everyone! Today, we're diving into a fascinating problem in real analysis that involves the convergence and divergence of infinite series. Specifically, we're exploring the existence of a positive-valued sequence (aₙ) such that the sum of (aₙ) raised to the power of r converges if and only if r is greater than or equal to 1. This is a classic problem that beautifully illustrates the interplay between the terms of a sequence and the behavior of its associated series. Let's break it down step by step.

The Challenge: Finding the Right Sequence

The core question we're tackling is this: Can we find a sequence of non-negative numbers, let's call it (aₙ), with a rather peculiar property? This property is that the infinite sum ∑(aₙ)ʳ converges (meaning it adds up to a finite value) precisely when r is greater than or equal to 1. For values of r less than 1, the sum should diverge, meaning it grows without bound. To kick things off, you might think about some familiar sequences and their convergence properties. For instance, the sequence aₙ = 1/n is a great starting point, as it's closely related to the well-known harmonic series and its variations. But does it fit our specific criteria? Let's investigate further.

Initial Thoughts and the Harmonic Series Connection

When first faced with this problem, it's natural to consider sequences related to the harmonic series. The harmonic series itself (∑ 1/n) is famous for its divergence, which means the sum of its terms grows infinitely large. However, if we raise the terms to a power greater than 1, like in the series ∑ 1/n², the series converges. This observation hints at a potential strategy. We need a sequence that behaves similarly but has a convergence threshold exactly at r = 1. The sequence aₙ = 1/n almost works, but it doesn't quite meet the 'if and only if' condition for r = 1. We need something a bit more refined. Think about how we can tweak the denominator to control the convergence more precisely. Maybe introducing logarithms or other functions could give us the desired behavior. This is where the real fun begins – the search for the perfect sequence!

Building Intuition: Why This is Tricky

Before diving into the solution, let's appreciate why this problem is interesting. It's not just about finding any sequence; it's about finding one with a very specific convergence behavior. The exponent r acts as a kind of dial, controlling whether the series converges or diverges. When r is small, the terms (aₙ)ʳ don't shrink fast enough, leading to divergence. When r is large, the terms shrink rapidly, guaranteeing convergence. The challenge is to find a sequence where the switch from divergence to convergence happens precisely at r = 1. This requires a delicate balance in the rate at which the terms aₙ decrease. We need them to decrease slowly enough that the series diverges for r < 1, but quickly enough that it converges for r ≥ 1. Understanding this trade-off is crucial for solving the problem.

Constructing the Sequence: A Logarithmic Approach

Okay, let's get down to brass tacks and construct the sequence. The key insight here is to use logarithms to fine-tune the convergence behavior. We'll define our sequence (aₙ) as follows:

aₙ = 1 / (n(log n)²) for n ≥ 2

Notice a few things about this sequence. First, it's positive for all n ≥ 2. Second, it decreases as n increases, which is essential for convergence analysis. The inclusion of the (log n)² term in the denominator is crucial. It makes the terms decrease slightly faster than 1/n, which, as we discussed earlier, is a key ingredient for achieving convergence when r = 1. But why this particular form? Let's see how it behaves when we raise it to the power of r and sum it up.

Analyzing Convergence for r = 1

Let's first consider the case when r = 1. We need to investigate the convergence of the series:

∑ (aₙ)¹ = ∑ 1 / (n(log n)²)

To determine convergence, we can use the integral test. The integral test is a powerful tool that connects the convergence of a series to the convergence of a corresponding integral. If the integral of a positive, decreasing function converges, then the series formed by sampling the function at integer points also converges, and vice versa. In our case, the function we're interested in is f(x) = 1 / (x(log x)²).

So, we need to evaluate the integral:

∫₂^∞ 1 / (x(log x)²) dx

To solve this integral, we can use a u-substitution. Let u = log x, so du = (1/x) dx. The limits of integration change accordingly: when x = 2, u = log 2, and as x approaches infinity, u also approaches infinity. The integral then becomes:

∫log₂^∞ 1 / u² du

This is a standard integral that evaluates to -1/u. Evaluating this from log 2 to infinity, we get:

[-1/u]log₂^∞ = (1/log 2)

Since the integral converges to a finite value (1/log 2), the integral test tells us that the series ∑ 1 / (n(log n)²) also converges. Great! Our sequence works for r = 1. Now, let's see what happens for r > 1.

Convergence for r > 1

Now, let's tackle the case where r is strictly greater than 1. We need to analyze the convergence of the series:

∑ (aₙ)ʳ = ∑ [1 / (n(log n)²)]ʳ = ∑ 1 / (nʳ(log n)²ʳ)

Here, the exponent r applies to both n and (log n)². To determine convergence, we can again use the integral test. The corresponding integral is:

∫₂^∞ 1 / (xʳ(log x)²ʳ) dx

This integral is a bit trickier than the previous one, but the key idea is that since r > 1, the xʳ term in the denominator will dominate the (log x)²ʳ term as x becomes large. In other words, for sufficiently large x, the function 1 / (xʳ(log x)²ʳ) behaves like 1/xʳ. We know that the integral ∫₁^∞ 1/xʳ dx converges if r > 1 (this is a standard result from p-series). Therefore, we can expect our integral to converge as well. A more rigorous proof would involve using comparison tests or more advanced integration techniques, but the intuition here is solid. The xʳ term ensures that the function decreases rapidly enough for the integral, and hence the series, to converge. So, our sequence also works for r > 1.

Divergence for r < 1: The Tricky Part

This is where things get interesting. We need to show that the series ∑ (aₙ)ʳ diverges when r is less than 1. This is often the most challenging part of these types of problems because we need to demonstrate that the terms don't shrink fast enough for the series to converge. Let's consider the series:

∑ (aₙ)ʳ = ∑ [1 / (n(log n)²)]ʳ = ∑ 1 / (nʳ(log n)²ʳ)

When r < 1, the exponent on n in the denominator is less than 1. This means that the terms 1/nʳ decrease more slowly than 1/n. However, we also have the (log n)²ʳ term, which decreases very slowly. The interplay between these two terms determines the convergence behavior. To show divergence, we'll use a comparison argument, but it needs to be done carefully. We can't directly compare it to 1/n because of the logarithmic term. Instead, we'll compare it to a slightly modified version of the original sequence.

The Comparison Strategy

The key insight is to recognize that for large enough n, (log n)²ʳ will be smaller than some power of n, say n^(ε), where ε is a small positive number. This is because logarithmic functions grow much more slowly than power functions. So, we can find an ε > 0 such that (log n)²ʳ < n^(ε) for sufficiently large n. Then, we have:

1 / (nʳ(log n)²ʳ) > 1 / (nʳn^(ε)) = 1 / n^(r+ε)

If we choose ε such that r + ε < 1, then the series ∑ 1 / n^(r+ε) will diverge (again, this is a property of p-series). By the comparison test, if ∑ 1 / n^(r+ε) diverges, then our original series ∑ 1 / (nʳ(log n)²ʳ) must also diverge. To make this rigorous, we need to choose ε carefully. Since r < 1, we can choose ε = (1 - r) / 2. Then, r + ε = r + (1 - r) / 2 = (1 + r) / 2, which is less than 1 because r < 1. This confirms our divergence result. Whew! That was a bit tricky, but we've shown that the series diverges for r < 1.

The Grand Finale: Putting It All Together

We've done it! We've constructed a sequence (aₙ) = 1 / (n(log n)²) that satisfies the given condition. Let's recap our findings:

• For r = 1, the series ∑ (aₙ)ʳ converges (by the integral test).
• For r > 1, the series ∑ (aₙ)ʳ converges (also by the integral test, and the dominance of the xʳ term).
• For r < 1, the series ∑ (aₙ)ʳ diverges (by a careful comparison argument).

Therefore, the series ∑ (aₙ)ʳ converges if and only if r ≥ 1. This completes our journey! We've successfully found a sequence with the desired convergence property. This problem highlights the power of combining different convergence tests and comparison arguments to analyze the behavior of infinite series. It also showcases the subtle interplay between different types of functions (polynomials and logarithms) in determining convergence.

Key Takeaways and Further Explorations

This problem is a great example of how seemingly simple questions in real analysis can lead to deep and interesting results. Here are a few key takeaways:

• The importance of logarithms: Logarithmic terms can be used to fine-tune the convergence behavior of series.
• The power of the integral test: The integral test is a powerful tool for relating series convergence to integral convergence.
• Comparison arguments: Careful comparison arguments are essential for proving divergence.

If you enjoyed this problem, here are a few avenues for further exploration:

• Can you find other sequences that satisfy the same condition?
• What happens if you replace (log n)² with a different function, like (log n)³ or log(log n)?
• Can you generalize this result to other types of series?

Real analysis is full of fascinating questions like this, and exploring them can deepen your understanding of the fundamental concepts of convergence and divergence. Keep exploring, keep questioning, and keep learning!

Conclusion: The Beauty of Convergence

Guys, we've successfully navigated a pretty complex problem, and hopefully, you've gained a deeper appreciation for the intricacies of convergence and divergence. This whole exercise wasn't just about finding a specific sequence; it was about honing our analytical skills and understanding the subtle dance between different mathematical concepts. Remember, real analysis is a playground for the mind, so keep those questions coming and never stop exploring! Happy analyzing!