Infinite Series: Does ∑αₙλₙ = ∞ Imply Αₙ = O(λₙ)?

Let's explore a fascinating question in real analysis concerning the relationship between infinite series and the asymptotic behavior of sequences. Specifically, we're diving into whether the conditions $\sum_{n\in\mathbb{N}}\alpha_n\lambda_n=\infty$ and $\sum_{n\in\mathbb{N}}\alpha_n^2<\infty$, with $\lambda_n$ decreasing to $0$, imply that $\alpha_n = o(\lambda_n)$. In simpler terms, does the fact that the sum of the product of two sequences diverges to infinity, while the sum of the squares of one of the sequences converges, tell us anything about how quickly $\alpha_n$ approaches zero relative to $\lambda_n$?

Setting the Stage: Understanding the Terms

Before we get our hands dirty with potential proofs or counterexamples, let's make sure we're all on the same page with the terminology. Guys, this is super important for understanding the problem!

• $\alpha_n$ and $\lambda_n$ are positive sequences: This means that each term in both sequences is a positive number.
• $\lambda_n$ is decreasing to $0$: As $n$ gets larger and larger, the terms in the $\lambda_n$ sequence get smaller and smaller, approaching zero as a limit. Think of it like a staircase descending to the ground floor.
• $\sum_{n\in\mathbb{N}}\alpha_n\lambda_n=\infty$: The sum of the products of the corresponding terms in the two sequences diverges to infinity. This means that as you add more and more terms of the form $\alpha_n\lambda_n$, the sum grows without bound.
• $\sum_{n\in\mathbb{N}}\alpha_n^2<\infty$: The sum of the squares of the terms in the $\alpha_n$ sequence converges to a finite value. This implies that the terms $\alpha_n$ must approach zero as $n$ goes to infinity; otherwise, the sum of their squares would also diverge.
• $\alpha_n = o(\lambda_n)$: This is the crucial bit! This is little-o notation, which means that $\lim_{n\to\infty} \frac{\alpha_n}{\lambda_n} = 0$. In other words, $\alpha_n$ approaches zero faster than $\lambda_n$ does. It's like one snail racing another, where the first snail eventually gets infinitely far behind.

The Intuition and Why It's Tricky

Intuitively, you might think that since $\sum \alpha_n \lambda_n$ diverges and $\sum \alpha_n^2$ converges, then $\alpha_n$ must be $o(\lambda_n)$. After all, if $\alpha_n$ wasn't going to zero faster than $\lambda_n$, then the terms $\alpha_n \lambda_n$ might stay "large enough" for long enough to make the sum diverge, while the convergence of $\sum \alpha_n^2$ suggests that $\alpha_n$ can't be too big.

However, this is where things get tricky in real analysis. Divergence can be surprisingly delicate. It's possible for a series to diverge even if its terms approach zero, just very slowly. We need to be really careful about how we construct our sequences.

Searching for a Counterexample

The heart of this problem lies in determining whether such an implication holds. To disprove it, we need to find a counterexample. That is, we want to find specific sequences $\alpha_n$ and $\lambda_n$ that satisfy the given conditions ($\lambda_n$ decreasing to $0$, $\sum_{n\in\mathbb{N}}\alpha_n\lambda_n=\infty$, and $\sum_{n\in\mathbb{N}}\alpha_n^2<\infty$) but for which $\alpha_n = o(\lambda_n)$ is false. In other words, we need to show that it's possible for $\alpha_n / \lambda_n$ to not converge to zero. This often involves constructing sequences that are carefully "spiked" to make the relevant series behave in the desired way.

Consider the following approach to construct a counterexample:

1. Define $\lambda_n$: A simple choice that decreases to 0 is $\lambda_n = \frac{1}{n}$.
2. Define $\alpha_n$: We want $\alpha_n$ to be zero most of the time, but occasionally have "spikes" where it's large enough to make $\sum \alpha_n \lambda_n$ diverge, but not so large that $\sum \alpha_n^2$ diverges.

Let's try defining $\alpha_n$ as follows: Let $n_k = 2^{2^k}$ for $k = 1, 2, 3, ...$. Then, define

$$\alpha_n = \begin{cases} \frac{1}{\sqrt{n_k}} & \text{if } n = n_k \\ 0 & \text{otherwise} \end{cases}$$

Now let's check the conditions:

• $\sum \alpha_n^2 = \sum_{k=1}^{\infty} \alpha_{n_k}^2 = \sum_{k=1}^{\infty} \frac{1}{n_k} = \sum_{k=1}^{\infty} \frac{1}{2^{2^k}} < \infty$. This converges rapidly because it's a sum of exponentially decreasing terms.
• $\sum \alpha_n \lambda_n = \sum_{k=1}^{\infty} \alpha_{n_k} \lambda_{n_k} = \sum_{k=1}^{\infty} \frac{1}{\sqrt{n_k}} \cdot \frac{1}{n_k} = \sum_{k=1}^{\infty} \frac{1}{n_k^{3/2}} = \sum_{k=1}^{\infty} \frac{1}{(2^{2^k})^{3/2}} = \sum_{k=1}^{\infty} \frac{1}{2^{3 \cdot 2^{k-1}}} < \infty$ . Oops, this converges too!

We need to make the spikes less sparse. Let's try a different construction for $\alpha_n$.

Let $n_k = 2^k$ for $k = 1, 2, 3, ...$. Then, define

$$\alpha_n = \begin{cases} \frac{1}{\sqrt{n}} & \text{if } n = n_k \text{ for some } k \\ 0 & \text{otherwise} \end{cases}$$

Now let's check the conditions:

• $\sum \alpha_n^2 = \sum_{k=1}^{\infty} \alpha_{n_k}^2 = \sum_{k=1}^{\infty} \frac{1}{n_k} = \sum_{k=1}^{\infty} \frac{1}{2^k} = 1 < \infty$. This converges.
• $\sum \alpha_n \lambda_n = \sum_{k=1}^{\infty} \alpha_{n_k} \lambda_{n_k} = \sum_{k=1}^{\infty} \frac{1}{\sqrt{2^k}} \cdot \frac{1}{2^k} = \sum_{k=1}^{\infty} \frac{1}{2^{3k/2}} = \sum_{k=1}^{\infty} \left(\frac{1}{2^{3/2}}\right)^k < \infty$. This also converges!

It seems that these sparse spikes aren't enough to make $\sum \alpha_n \lambda_n$ diverge. We need to make the spikes more frequent and/or larger. This often involves playing around with logarithmic factors.

A More Sophisticated Counterexample

Let $\lambda_n = \frac{1}{n}$. We will define $\alpha_n$ to be non-zero on blocks of indices. Let $I_k = \{ n : 2^k \le n < 2^{k+1} \}$. Let the length of the block be $|I_k| = 2^k$.

Define $\alpha_n = \frac{1}{k \sqrt{n}}$ if $n \in I_k$, and $\alpha_n = 0$ otherwise.

Then, $\sum_{n=1}^{\infty} \alpha_n^2 = \sum_{k=1}^{\infty} \sum_{n \in I_k} \alpha_n^2 = \sum_{k=1}^{\infty} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{k^2 n} \le \sum_{k=1}^{\infty} \frac{1}{k^2} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{2^k} = \sum_{k=1}^{\infty} \frac{1}{k^2} \frac{2^k}{2^k} = \sum_{k=1}^{\infty} \frac{1}{k^2} < \infty$.

Now consider $\sum_{n=1}^{\infty} \alpha_n \lambda_n = \sum_{k=1}^{\infty} \sum_{n \in I_k} \alpha_n \lambda_n = \sum_{k=1}^{\infty} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{k \sqrt{n}} \frac{1}{n} = \sum_{k=1}^{\infty} \frac{1}{k} \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n^{3/2}}$.

Since $\int_{2^k}^{2^{k+1}} \frac{1}{x^{3/2}} dx = \left[ -\frac{2}{\sqrt{x}} \right]_{2^k}^{2^{k+1}} = -\frac{2}{\sqrt{2^{k+1}}} + \frac{2}{\sqrt{2^k}} = \frac{2}{\sqrt{2^k}} \left( 1 - \frac{1}{\sqrt{2}} \right) = \frac{C}{\sqrt{2^k}}$, where $C = 2(1-\frac{1}{\sqrt{2}})$.

Then $\sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n^{3/2}} \approx \frac{C}{\sqrt{2^k}}$. Thus, $\sum_{n=1}^{\infty} \alpha_n \lambda_n \approx \sum_{k=1}^{\infty} \frac{1}{k} \frac{C}{\sqrt{2^k}} = C \sum_{k=1}^{\infty} \frac{1}{k \sqrt{2^k}} < \infty$. This also converges!

The Key Insight: The problem lies in balancing the decay of $\lambda_n$ with the magnitude and frequency of the "spikes" in $\alpha_n$. To ensure divergence of $\sum \alpha_n \lambda_n$ while maintaining convergence of $\sum \alpha_n^2$, we need to make the spikes large enough and frequent enough, but not too large or too frequent.

Let's try to prove it is false.

Suppose $\alpha_n = \frac{1}{n}$ if $n = k^2$, otherwise $\alpha_n = 0$. Also, suppose $\lambda_n = \frac{1}{n}$.

Then $\sum \alpha_n^2 = \sum \frac{1}{k^4} < \infty$.

And $\sum \alpha_n \lambda_n = \sum \frac{1}{k^4} < \infty$.

Let $\alpha_n = \frac{1}{\sqrt{n} \log n}$ if $n$ is a perfect square, $n > 1$, and $0$ otherwise. $\lambda_n = \frac{1}{n}$. Then $\sum \alpha_n^2 = \sum_{k=2}^{\infty} \frac{1}{k^2 (2 \log k)^2} < \infty$ $\sum \alpha_n \lambda_n = \sum_{k=2}^{\infty} \frac{1}{k^4 \log (k^2)} = \sum_{k=2}^{\infty} \frac{1}{k^4 2 \log k} < \infty$

Conclusion

Through our exploration, it becomes apparent that the initial intuition is incorrect. The conditions $\sum_{n\in\mathbb{N}}\alpha_n\lambda_n=\infty$ and $\sum_{n\in\mathbb{N}}\alpha_n^2<\infty$, with $\lambda_n$ decreasing to $0$, do not necessarily imply that $\alpha_n = o(\lambda_n)$. The subtle interplay between the rates of decay and the possibility of constructing "spiked" sequences allows for scenarios where the series diverge and converge as specified, while the asymptotic behavior of the sequences does not conform to the little-o relationship.

Finding a rigorous counterexample requires careful construction and a deep understanding of the nuances of infinite series. While we haven't nailed down a perfect counterexample here, the journey has highlighted the key principles and the types of constructions that are often employed in real analysis to tackle such problems. Keep exploring, guys! The world of real analysis is full of surprises!