What is the value of \(1 + \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{64}} + \frac{1}{{256}} + \ldots ?\)
Value of an Infinite Geometric Series
The question asks us to find the value of the infinite series: \(1 + \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{64}} + \frac{1}{{256}} + \ldots\). This type of series, where each term after the first is found by multiplying the previous one by a fixed, non-zero number, is known as a geometric series.
Since the series continues indefinitely (indicated by the "..." at the end), it is an infinite geometric series. To find the sum of an infinite geometric series, we need to identify its first term and its common ratio.
Identifying Series Components
Let's break down the given infinite series to identify its key components:
- First Term (\(a\)): The first term in the series is \(1\). So, \(a = 1\).
- Common Ratio (\(r\)): The common ratio is found by dividing any term by its preceding term. Let's pick the second term and divide it by the first term: \[r = \frac{\frac{1}{4}}{1} = \frac{1}{4}\] We can verify this by taking other consecutive terms: \[r = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{16} \times 4 = \frac{4}{16} = \frac{1}{4}\] Since the ratio is consistent, the common ratio \(r = \frac{1}{4}\).
Summing an Infinite Geometric Series
An infinite geometric series converges to a finite sum if and only if the absolute value of its common ratio (\(r\)) is less than 1 (i.e., \(|r| < 1\)). In this case, \(r = \frac{1}{4}\), and \(|\frac{1}{4}| = \frac{1}{4}\), which is indeed less than 1. Therefore, the series converges, and we can find its sum.
The formula for the sum \(S\) of a convergent infinite geometric series is given by:
\[S = \frac{a}{1 - r}\]
Now, let's substitute the values of \(a\) and \(r\) that we found:
- \(a = 1\)
- \(r = \frac{1}{4}\)
Plugging these values into the formula:
\[S = \frac{1}{1 - \frac{1}{4}}\]
First, calculate the denominator:
\[1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{4 - 1}{4} = \frac{3}{4}\]
Now, substitute this back into the sum formula:
\[S = \frac{1}{\frac{3}{4}}\]
To divide by a fraction, we multiply by its reciprocal:
\[S = 1 \times \frac{4}{3}\]
\[S = \frac{4}{3}\]
Final Value
The value or sum of the given infinite series \(1 + \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{64}} + \frac{1}{{256}} + \ldots\) is \(\frac{4}{3}\).
| Component | Value |
|---|---|
| First Term (\(a\)) | \(1\) |
| Common Ratio (\(r\)) | \(\frac{1}{4}\) |
| Sum of Infinite Geometric Series (\(S\)) | \(\frac{4}{3}\) |
Important Questions from Numerical Computation
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