Question

determine whether the series is convergent or divergent. if it is convergent, find its sum. (if the quantity diverges, enter diverges.)

\\\frac{1}{2} + \frac{3}{4} + \frac{1}{8} + \frac{3}{16} + \frac{1}{32} + \frac{3}{64} + \cdots\\

Explanation:

⚡ Using what you learned: Tests for Convergence (nth-term, comparison, ratio, root, alternating) · 🆕 New Concept Discovered: Geometric Series Sum
Summing infinite terms with a constant ratio

Step 1: Identify the pattern and split the series

The given series is:

$$ S = \frac{1}{2} + \frac{3}{4} + \frac{1}{8} + \frac{3}{16} + \frac{1}{32} + \frac{3}{64} + \cdots $$

We can split this series into two separate infinite series by grouping the odd-positioned terms and the even-positioned terms:

$$ S_1 = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots $$

$$ S_2 = \frac{3}{4} + \frac{3}{16} + \frac{3}{64} + \cdots $$

Step 2: Sum the first geometric series

For \( S_1 \):

• First term \( a_1 = \frac{1}{2} \)
• Common ratio \( r_1 = \frac{1}{4} \)

Since \( |r_1| < 1 \), the series converges. Its sum is:

$$ S_1 = \frac{a_1}{1 - r_1} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3} $$

Step 3: Sum the second geometric series

For \( S_2 \):

• First term \( a_2 = \frac{3}{4} \)
• Common ratio \( r_2 = \frac{1}{4} \)

Since \( |r_2| < 1 \), the series converges. Its sum is:

$$ S_2 = \frac{a_2}{1 - r_2} = \frac{\frac{3}{4}}{1 - \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4}} = 1 $$

Step 4: Combine the sums

Since both component series converge absolutely, the sum of the original series is the sum of their individual limits:

$$ S = S_1 + S_2 = \frac{2}{3} + 1 = \frac{5}{3} $$

Answer:

$$ \frac{5}{3} $$