We are tasked with expressing the sum $\frac{5}{3} + \frac{5}{6} + \frac{5}{12} + \frac{5}{24} + \dots$ as a series of the form $\sum_{n=0}^{\infty} a_n$ or $\sum_{n=1}^{\infty} a_n$, and then determining if the series converges.

Step 1: Identify the pattern

We observe that the denominators of the terms form a sequence: $3, 6, 12, 24, \dots$. This looks like a geometric progression where each denominator is doubled in the subsequent term.

• The first term is $\frac{5}{3}$.
• The second term is $\frac{5}{6} = \frac{5}{3 \cdot 2}$.
• The third term is $\frac{5}{12} = \frac{5}{3 \cdot 2^2}$.
• The fourth term is $\frac{5}{24} = \frac{5}{3 \cdot 2^3}$.

Hence, the general term for this series can be written as: $a_n = \frac{5}{3 \cdot 2^n}, \quad n \geq 0$

Thus, the series can be expressed as: $\sum_{n=0}^{\infty} \frac{5}{3 \cdot 2^n}$

Step 2: Simplify and find the sum of the series

This is a geometric series of the form: $\sum_{n=0}^{\infty} \frac{5}{3} \cdot \left( \frac{1}{2} \right)^n$

The sum of an infinite geometric series $\sum_{n=0}^{\infty} ar^n$ is given by: $S = \frac{a}{1 - r} \quad \text{if} \quad |r| < 1$

Here, $a = \frac{5}{3}$ and $r = \frac{1}{2}$. Therefore, the sum of the series is: $S = \frac{\frac{5}{3}}{1 - \frac{1}{2}} = \frac{\frac{5}{3}}{\frac{1}{2}} = \frac{5}{3} \cdot 2 = \frac{10}{3}$

Conclusion:

The series converges, and the sum is: $S = \frac{10}{3}$

Would you like further details or have any questions?

Here are 5 related questions for further practice:

1. What conditions are necessary for a geometric series to converge?
2. How do you determine the common ratio in a geometric series?
3. What is the formula for the sum of a geometric series with a given first term and ratio?
4. If the first term of the series was $\frac{7}{4}$ instead of $\frac{5}{3}$, how would the sum change?
5. Can you derive the sum of a geometric series where the ratio is negative?

Tip: When analyzing infinite series, always check whether the common ratio lies within the interval $(-1, 1)$ for the series to converge.