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The comparison test is a powerful tool used to determine the convergence or divergence of a series. When applying the comparison test, we compare the series in question to another series whose convergence properties we already understand.

To apply this test, consider two series:

• The given series: \( \sum a_k \)
• A known series: \( \sum b_k \)

If \( 0 \leq a_k \leq b_k \) for all terms, and \( \sum b_k \) converges, then \( \sum a_k \) also converges. Conversely, if \( a_k \geq b_k \) for all terms, and \( \sum b_k \) diverges, then \( \sum a_k \) diverges as well.

In our specific example, once we establish that \( \frac{1}{k 2^k} \leq \frac{1}{2^k} \), and we know that the geometric series \( \sum \frac{1}{2^k} \) converges, we can conclude that the original series \( \sum \frac{1}{k 2^k} \) converges using the comparison test.

A geometric series is a series of the form \( \sum ar^{k} \), where \( a \) is the initial term and \( r \) is the common ratio. The series \( \sum \frac{1}{2^k} \) is a classic example of a geometric series with \( a = 1 \) and \( r = \frac{1}{2} \).

Geometric series are particularly simple to evaluate for convergence. The series converges to \( \frac{a}{1-r} \) if \( |r| < 1 \). If \( |r| \geq 1 \), the series diverges.

In our exercise, the series \( \sum \frac{1}{2^k} \) has \( r = \frac{1}{2} \), which is less than 1, so it converges. This makes it a suitable series for comparison when applying the comparison test to the original series \( \sum \frac{1}{k 2^k} \).

To use the comparison test effectively, demonstrating an inequality between the series terms is crucial. Here, we show that each term of the original series \( \frac{1}{k 2^k} \) is dominated by the corresponding term of the geometric series \( \frac{1}{2^k} \).

The inequality \( \frac{1}{k 2^k} \leq \frac{1}{2^k} \) simplifies to \( k \geq 1 \), which always holds true since \( k \) is a positive integer starting from 1. This means every term in the original series is smaller than its counterpart in the convergent geometric series.

Establishing such inequalities ensures that the comparison test can be applied successfully. Therefore, since the larger series converges, the smaller one must converge as well.