Question

Prove: If \(\sum\left|a_{n}\right|<\infty\) and \(\sum b_{n}\) converges (perhaps conditionally), with \(\sum_{n=0}^{\infty} a_{n}=\) \(A\) and \(\sum_{n=0}^{\infty} b_{n}=B,\) then the Cauchy product $$ \sum_{n=0}^{\infty} c_{n}=\left(\sum_{n=0}^{\infty} a_{n}\right)\left(\sum_{n=0}^{\infty} b_{n}\right) $$ converges to \(A B\). HINT: Let \(\left\\{A_{n}\right\\},\left\\{B_{n}\right\\}\), and \(\left\\{C_{n}\right\\}\) be the partial sums of the series. Show that $$ C_{n}-A_{n} B=\sum_{r=0}^{n} a_{r}\left(B_{n-r}-B\right) $$ and apply Theorem 4.3 .5 to \(\sum\left|a_{n}\right|\)

Short answer

Question: Prove that if the series \(\sum\left|a_{n}\right|\) converges and \(\sum b_{n}\) converges (possibly conditionally), then the Cauchy product \(\sum_{n=0}^{\infty} c_{n}\) converges and equals the product of the sums of the two series (i.e., \(\sum_{n=0}^{\infty} c_{n} = \lim_{n\to\infty} A_n \lim_{n\to\infty} B_n\)). Hint: Define the partial sums \(A_n = \sum_{i=0}^n a_i\), \(B_n = \sum_{i=0}^n b_i\), and \(C_n = \sum_{i=0}^n c_i\). Write down an expression for \(C_n - A_n B\) in terms of the given sequences, and apply the Absolute Convergence Test (Theorem 4.3.5).

Step-by-step solution

Define the partial sums

Let \(A_n = \sum_{i=0}^n a_i\), \(B_n = \sum_{i=0}^n b_i\), and \(C_n = \sum_{i=0}^n c_i\). Then, \(\lim_{n\to\infty} A_n = A\), \(\lim_{n\to\infty} B_n = B\), and we want to show that \(\lim_{n\to\infty} C_n = AB\).

Express the difference \(C_n - A_n B\) in terms of partial sums

We want to find an expression for \(C_n - A_n B\) in terms of the given series. First, note that $$ c_n = \sum_{r=0}^n a_r b_{n-r}. $$ Then, $$ C_n = \sum_{k=0}^n c_k = \sum_{k=0}^n \sum_{r=0}^k a_r b_{k-r}, $$ and $$ C_n - A_n B = \sum_{k=0}^n \sum_{r=0}^k a_r b_{k-r} - \sum_{r=0}^n a_r \sum_{k=0}^n b_{k-r}. $$ By rearranging the terms and using the distributive property, we get $$ C_n - A_n B = \sum_{r=0}^n a_r \left(\sum_{k=0}^n b_{k-r} - \sum_{k=0}^n b_k\right) = \sum_{r=0}^n a_r(B_{n-r} - B). $$

Apply the Absolute Convergence Test (Theorem 4.3.5)

Since \(\sum |a_n|\) converges, it satisfies the Absolute Convergence Test. This means that $$ \lim_{n\to\infty} \sum_{r=0}^n |a_r||B_{n-r} - B| = 0. $$ Now, recall the following inequality for real numbers: \(|ab| \le |a||b|\). Using this inequality, we have $$ \left|\sum_{r=0}^n a_r(B_{n-r} - B)\right| \le \sum_{r=0}^n |a_r||B_{n-r} - B|, $$ and since the right-hand side converges to 0, we also have that $$ \lim_{n\to\infty} \left|\sum_{r=0}^n a_r(B_{n-r} - B)\right| = 0. $$ But this means that $$ \lim_{n\to\infty} \left|C_n - A_n B\right| = 0, $$ which implies $$ \lim_{n\to\infty} C_n = \lim_{n\to\infty} (A_n B) = AB. $$ Thus, we have proved that the Cauchy product converges to the product of the sums, i.e., \(\sum_{n=0}^{\infty} c_{n} = AB\).

Key concepts

Absolute Convergence

When we say a series converges absolutely, we mean that the series of the absolute values of its terms also converges. For example, if the series \(\sum a_n\) converges absolutely, then \(\sum |a_n|\) also converges. Absolute convergence is a stronger condition than regular convergence. This is because if a series converges absolutely, it also converges (but not necessarily the other way around).
This property is particularly useful when dealing with the Cauchy product of two series. If \(\sum |a_n| < \infty\), it ensures that we can rearrange the terms of the series without affecting convergence. This flexibility is crucial when dealing with complex series operations, such as those involved in the Cauchy product. Applying a transformation, like taking partial sums, becomes more straightforward with absolute convergence. It allows us to maintain control over the convergence behavior as we manipulate the series.

Partial Sums

Partial sums represent the sum of the first \(n\) terms of a series. For a given series \(\sum a_n\), the partial sum is represented as \(A_n = \sum_{i=0}^n a_i\).
These sums help us analyze the behaviour and convergence of a series by providing a sequence of progressively larger sums. By observing \(A_n\), we aim to see whether it approaches a finite limit as \(n\) goes to infinity. In the context of proving the Cauchy product for two convergent series \(\sum a_n\) and \(\sum b_n\), partial sums \(A_n\) and \(B_n\) are invaluable tools. They help demonstrate the relationship between the original series and the resulting series from their product. As in our exercise, \(C_n\) represents the partial sum of the resulting Cauchy product series. To show that \(C_n\) approaches \(AB\) (where \(A\) and \(B\) are the limits of the partial sums of the original series), we break down the expression using such partial sums.

Series Convergence

Series convergence is concerned with whether the sum of an infinite series approaches a specific value as the number of terms grows. For a series \(\sum a_n\) to converge to a limit \(A\), the sequence of partial sums \(A_n = \sum_{i=0}^n a_i\) must approach \(A\) as \(n\) tends toward infinity.
A series is said to be convergent if this condition holds true; it diverges if it does not.

• Convergent Series: These series reach a finite limit. Examples include geometric series with a ratio less than one.
• Divergent Series: These series do not reach a specific limit, but instead grow indefinitely or oscillate.

Applying these principles to our exercise, we examine the convergence of \(\sum b_n \) to \(B\). When analyzing the Cauchy product, knowing that the original series \(\sum a_n\) has terms \(a_n\) which are absolutely convergent, aids in proving the convergence to the product \(AB\), guaranteeing a well-defined convergence behavior for \(\sum c_n \).