Question

Suppose that \(a_{r} \geq 0\) for all \(r \geq 0\) and and \(\sum_{0}^{\infty} a_{r}=A<\infty .\) Show that $$ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s}=0 \quad \text { and } \quad \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s}=2 A-a_{0} $$

Short answer

**Question**: Prove the following limit statements for sequence \(a_r\) where \(a_r \geq 0\) for all \(r \geq 0\), and \(\sum_{0}^{\infty} a_{r} = A < \infty\): 1. \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s} = 0\) 2. \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} = 2A - a_0\) **Answer**: By analyzing the double summations involved in both limit statements, and employing simplification and use of the Cauchy-Schwarz Inequality in the first statement, we can show that: 1. \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s} = 0\) 2. \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} = 2A - a_0\)

Step-by-step solution

First Limit: Analyze the double summation in the first statement

To prove the first statement, start by observing the double summation \(\frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s}\) which represents the sum of all the terms \(a_{r+s}\) with \(r\) and \(s\) going from \(0\) to \(n-1\).

First Limit: Partially compute the double summation in the first statement

Next, observe that there are \(n(n-1)/2\) distinct values of \(r+s\) in that sum, since it counts all combinations from \((0,0)\) to \((n-1,n-1)\). Therefore, we can rewrite the double summation as follows: $$ \frac{1}{n} \sum_{t=0}^{2(n-1)} c(t), $$ where $$ c(t) = \sum_{r+s=t} a_{r+s}. $$

First Limit: Apply Cauchy-Schwarz Inequality

To evaluate the limit, use the Cauchy-Schwarz Inequality: $$ \left(\frac{1}{n} \sum_{t=0}^{2(n-1)} c(t)\right)^2 \le \left(\frac{1}{n} \sum_{t=0}^{2(n-1)} a_{t}\right) \left(\frac{1}{n} \sum_{t=0}^{2(n-1)} a_{t}\right), $$ which simplifies to: $$ \left(\frac{1}{n} \sum_{t=0}^{2(n-1)} c(t)\right)^2 \le \left(\frac{1}{n} \sum_{t=0}^{2(n-1)} a_{t}\right)^2. $$

First Limit: Compute the limit

Now, observe that since the series \(\sum_{0}^{\infty} a_{r}\) converges, we have: $$ \lim_{n \to \infty} \frac{1}{n} \sum_{t=0}^{2(n-1)} a_{t} = 0. $$ Hence, by applying the limit to both sides of the inequality obtained in Step 3, we get the desired result: $$ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s} = 0. $$

Second Limit: Analyze the double summation in the second statement

To prove the second statement, start by observing the double summation \(\frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s}\) which represents the sum of all the terms \(a_{r-s}\) with \(r\) and \(s\) going from \(0\) to \(n-1\).

Second Limit: Compute some cases for the second summation

Compute some cases of the second summation: - For \(r = s\), the terms \(a_{r-s} = a_0\), giving the sum \(na_0\), - For \(r < s\), the terms \(a_{r-s}\) include \(a_1, a_2, ...a_{n-1}\), giving the sum \(a_1 + a_2 + ... + a_{n-1} = A - a_0 - a_n - a_{n+1} - ...\), - For \(r > s\), the terms \(a_{r-s}\) include \(a_{1}, a_{2}, ... a_{n-1}\), giving the sum \(A-a_0-a_n-a_{n+1}-...\).

Second Limit: Add the cases for the second summation

Adding these cases together, and remembering to divide by \(n\) (as in the definition), we get: $$ \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} = \frac{na_{0} + 2(A-a_{0}-a_{n}-a_{n+1}-...)}{n}. $$

Second Limit: Compute the limit

To find the limit as \(n\) approaches infinity, we first notice that \(a_n, a_{n+1}, ...\) all approach zero since the series is convergent. Thus, we have: $$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} = \lim_{n \rightarrow \infty} \frac{na_{0} + 2(A-a_{0}-a_{n}-a_{n+1}-...)}{n}. $$ Which yields the desired result: $$ \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} = 2A - a_0. $$

Key concepts

Convergent Series

When we speak of a convergent series, we're referring to a series of numbers that, when added together, approach a certain value as more and more terms are included. This concept is incredibly important in real analysis as it helps to understand the behavior of infinite sequences and series.

In our context, the given series \( \sum_{0}^{\infty} a_{r} \) has a finite sum, denoted as \( A \). This convergence implies that as \( r \) increases, the terms \( a_{r} \) get smaller and eventually tend to zero, which is a central point for analyzing the limits of sequences involving these terms.

Understanding convergent series allows us to make predictions about the behavior of sequences over the long term, and this has applications not only in pure mathematics but in fields like physics and engineering where modeling infinite processes is essential.

Double Summation

A double summation is a form of multiple summation whereby two indices are summed over in a nested fashion. It's common in mathematics when dealing with two-dimensional data or bivariate functions. In our exercise, the double summation appears in the form \( \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r+s} \) and \( \frac{1}{n} \sum_{r, s=0}^{n-1} a_{r-s} \).

To address the complexity of a double summation, one can often rearrange or break down the summation into simpler parts that can be more easily computed or analyzed. For example, in the first limit of our exercise, the double summation is partially computed by recognizing and rearranging the values into a single summation over a new index \( t \) that combines \( r \) and \( s \). This is a powerful technique that simplifies the evaluation of the limit.

Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is a fundamental result in linear algebra and real analysis. It provides an upper bound on the product of two vectors' dot product and is expressed as:

\[\left( \sum_{i} a_{i}b_{i} \right)^2 \leq \left( \sum_{i} a_{i}^2 \right) \left( \sum_{i} b_{i}^2 \right)\]

In terms of series, this inequality suggests a bound on the sum of products of terms from two series by the product of the sums of squares of the individual series. In the exercise, the inequality is creatively applied to the terms of the convergent series to find an upper bound, which greatly facilitates the computation of the limit. The Cauchy-Schwarz Inequality is a powerful tool in proving various results involving limits and convergence in real analysis.

Limit of a Sequence

The limit of a sequence is the value that the elements of a numerical sequence get closer to as the sequence progresses. Mathematically, we say that the sequence \( a_n \) has a limit \( L \) if for any small positive number \( \varepsilon \) we choose, there exists a corresponding positive integer \( N \) such that for all \( n > N \) the distance between \( a_n \) and \( L \) is less than \( \varepsilon \)—in mathematical terms, \( |a_n - L| < \varepsilon \).

In our problem, we deal with the limits of sequences expressed as sums. We needed to demonstrate that two specific sequences of sums approached certain values as \( n \) went to infinity. The ability to accurately determine the limit of sequences is a cornerstone in the analysis of mathematical series, functions, and various real-world phenomena modeled by sequences and series.