Problem 1 - A Course in Calculus and Real Analysis

Question

Give examples to show that if \(\sum_{k} a_{k}\) and \(\sum_{k} b_{k}\) are convergent series of real numbers, then the series \(\sum_{k} a_{k} b_{k}\) may not be convergent. Also show that if \(\sum_{k} a_{k}=A\) and \(\sum_{k} b_{k}=B\), then \(\sum_{k} a_{k} b_{k}\) may be convergent, but its sum may not be equal to \(A B\).

Step-by-Step Solution

Answer

Example 1: Choose \(a_{k}=(-1)^k\frac{1}{\sqrt{k}}\) and \(b_{k}=(-1)^{k+1}\frac{1}{\sqrt{k}}\). Both series \(\sum_{k} a_{k}\) and \(\sum_{k} b_{k}\) are convergent. However, the product series \(\sum_{k} a_{k} b_{k}\) has terms \(\frac{1}{k}\), which form a divergent harmonic series. Example 2: Choose \(a_{k}=\frac{1}{2^k}\) and \(b_{k}=(-1)^{k}\frac{1}{2^k}\). The series \(\sum_{k} a_{k}\) converges to 1 and \(\sum_{k} b_{k}\) converges to \(-\frac{1}{3}\). The product series \(\sum_{k} a_{k} b_{k}\) converges to \(-\frac{1}{5}\). In this case, the sum of the product series is not equal to the product of the sums, which is \(-\frac{1}{3}\).

Step 1: Choose \(a_k\) and \(b_k\) Series

Let \(a_{k}=(-1)^k\frac{1}{\sqrt{k}}\) and \(b_{k}=(-1)^{k+1}\frac{1}{\sqrt{k}}\).

Step 2: Convergence of \(a_k\) and \(b_k\) Series

Both series \(\sum_{k} a_{k}\) and \(\sum_{k} b_{k}\) are convergent by the Alternating Series Test, because the terms \(\frac{1}{\sqrt{k}}\) are decreasing and have the limit 0 when \(k \rightarrow \infty\).

Step 3: Divergence of the \(a_k b_k\) Series

Consider the product series \(\sum_{k} a_{k} b_{k}\). The terms of this series are given by \[a_k b_k = (-1)^k\frac{1}{\sqrt{k}}(-1)^{k+1}\frac{1}{\sqrt{k}} = \frac{1}{k}.\] So, the series \(\sum_{k} a_{k} b_{k}\) is a harmonic series: \[\sum_{k=1}^\infty \frac{1}{k}.\] This series is well-known to be divergent. Hence, we have a situation where \(\sum_{k} a_{k}\) and \(\sum_{k} b_{k}\) are both convergent, but \(\sum_{k} a_{k} b_{k}\) is divergent. #Example 2: Convergent \(\sum_{k} a_{k} b_{k}\) but not equal to \(AB\)#

Step 4: Choose \(a_k\) and \(b_k\) Series

Let \(a_{k}=\frac{1}{2^k}\) and \(b_{k}=(-1)^{k}\frac{1}{2^k}\).

Step 5: Convergence and Values of \(a_k\) and \(b_k\) Series

Note that \(\sum_{k} a_{k}\) is a geometric series with common ratio \(\frac{1}{2}\), and it converges to \[\sum_{k=1}^\infty \frac{1}{2^k} = \frac{1}{1-\frac{1}{2}} - 1 = 1.\] Similarly, \(\sum_{k} b_{k}\) is an alternating geometric series with common ratio \(-\frac{1}{2}\) and it converges to \[\sum_{k=1}^\infty (-1)^{k}\frac{1}{2^k} = -\frac{1}{1+\frac{1}{2}} + 1 = -\frac{1}{3}.\]

Step 6: Convergence of the \(a_k b_k\) Series and Its Sum

The product series \(\sum_{k} a_{k} b_{k}\) has terms \[a_k b_k = \frac{1}{2^k}(-1)^{k}\frac{1}{2^k} = \frac{(-1)^{k}}{4^k}.\] This series is an alternating geometric series with common ratio \(\frac{1}{4}\), which converges to \[\sum_{k=1}^\infty \frac{(-1)^{k}}{4^k} = -\frac{1}{1+\frac{1}{4}} + 1 = -\frac{1}{5}.\]

Step 7: Sum of Product Series not Equal to \(AB\)

We found that \(\sum_{k} a_{k} = 1\), \(\sum_{k} b_{k} = -\frac{1}{3}\), and \(\sum_{k} a_{k} b_{k} = -\frac{1}{5}\). The product of the sums of the \(a_k\) and \(b_k\) series is,\(AB = (1)\left(-\frac{1}{3}\right) = -\frac{1}{3}.\) However, the sum of the product series \(\sum_{k} a_{k} b_{k}\) is \(-\frac{1}{5}\), not equal to \(AB\).