Infinite Series Convergence Analysis

The infinite series is given as $1 + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$.

Let's examine the terms:

• First term: $T_1 = 1$
• Second term: $T_2 = \frac{1}{4}$
• Third term: $T_3 = \frac{1}{8}$
• Fourth term: $T_4 = \frac{1}{16}$

Calculate the ratios between consecutive terms:

• Ratio between $T_2$ and $T_1$: $\frac{T_2}{T_1} = \frac{1/4}{1} = \frac{1}{4}$
• Ratio between $T_3$ and $T_2$: $\frac{T_3}{T_2} = \frac{1/8}{1/4} = \frac{1}{8} \times \frac{4}{1} = \frac{1}{2}$
• Ratio between $T_4$ and $T_3$: $\frac{T_4}{T_3} = \frac{1/16}{1/8} = \frac{1}{16} \times \frac{8}{1} = \frac{1}{2}$

From the second term onwards, the series behaves like a geometric progression: $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$.

This geometric sub-series has:

• First term $a = \frac{1}{4}$
• Common ratio $r = \frac{1}{2}$

Convergence Condition for Geometric Series

A geometric series converges if the absolute value of its common ratio is less than 1, i.e., $|r| < 1$.

For the sub-series, $|r| = |\frac{1}{2}| = \frac{1}{2}$.

Since $\frac{1}{2} < 1$, the geometric sub-series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ converges.

The sum of this convergent sub-series is calculated as $S_{sub} = \frac{a}{1-r} = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.

The original series is the sum of the first term and the convergent sub-series: $S = 1 + S_{sub} = 1 + \frac{1}{2} = \frac{3}{2}$.

Because the series sums to a finite value ($\frac{3}{2}$), the series is convergent.