Series Analysis: Structure

The given infinite series is: $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$

The series comprises a constant term and terms derived from powers of prime numbers.

• Constant Term: $2$
• Powers of 2 Series: The terms $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$ form a geometric series $\sum_{k=1}^{\infty} \frac{1}{2^k}$.
• Powers of 3 Series: The terms $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ form a geometric series $\sum_{k=1}^{\infty} \frac{1}{3^k}$.

The total sum (S) is expressed as: $S = 2 + \sum_{k=1}^{\infty} \frac{1}{2^k} + \sum_{k=1}^{\infty} \frac{1}{3^k}$

Geometric Series: Sum Calculation

We use the formula for the sum of an infinite geometric series, $S_{\infty} = \frac{a}{1-r}$, where '$a$' is the first term and '$r$' is the common ratio ($|r| < 1$).

Powers of 2 Series: Sum

For the series $\sum_{k=1}^{\infty} \frac{1}{2^k}$: First term $a = \frac{1}{2}$. Common ratio $r = \frac{1}{2}$. Sum$_2 = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1$.

Powers of 3 Series: Sum

For the series $\sum_{k=1}^{\infty} \frac{1}{3^k}$: First term $a = \frac{1}{3}$. Common ratio $r = \frac{1}{3}$. Sum$_3 = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$.

Total Series: Sum Calculation

Combine the constant term and the calculated sums:

Total Sum = $2 + \text{Sum}_2 + \text{Sum}_3$ Total Sum = $2 + 1 + \frac{1}{2}$ Total Sum = $3 + \frac{1}{2}$ Total Sum = $\frac{6}{2} + \frac{1}{2} = \frac{7}{2}$