Problem 36 - Algebra and Trigonometry

Question

Find the indicated sum. Use the formula for the sum of the first n terms of a geometric sequence. $$ \sum_{i=1}^{6}\left(\frac{1}{3}\right)^{i+1} $$

Step-by-Step Solution

Answer

The sum of the geometric series \( \sum_{i=1}^{6}(1/3)^{i+1} \) is approximately 0.13168.

Step 1: Identify the first term, common ratio, and number of terms

In this problem, the series is \( \sum_{i=1}^{6}(1/3)^{i+1} \). The first term can be found by substituting i=1 into the formula, giving \( a = (1/3)^{1+1} = 1/9 \). The ratio is the constant multiplying factor, which is \( r = 1/3 \). The number of terms is given by the upper limit on the summation, which is \( n = 6 \).

Step 2: Substitute into the sum formula

Substitute the values of \( a \), \( r \), and \( n \) into the sum formula for a geometric series \( S = a (1-r^n)/(1-r) \). So, \( S = (1/9) (1-(1/3)^6) / (1-1/3) \)

Step 3: Simplify the expression

Simplify the above expression to find the sum of the series. Therefore, \( S = 0.13168 \).

Key Concepts

Sum of Geometric Series

Understanding the sum of a geometric series can be quite straightforward with the right approach. A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. To find the sum of a series like \( \sum_{i=1}^{6}\left(\frac{1}{3}\right)^{i+1} \), you’ll need to apply a specific formula: \[ S_n = \frac{a(1-r^n)}{1-r} \], where \( S_n \) is the sum of the first \( n \) terms of the sequence, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

For our example, by identifying the first term \( a = \frac{1}{9} \), the common ratio \( r = \frac{1}{3} \), and the number of terms \( n = 6 \), we can plug these into the formula to get \( S_n = \frac{(1/9)(1-(1/3)^6)}{1-1/3} \). By simplifying this expression, we reach the series sum, \( S_n = 0.13168 \). This sum helps us understand how finite series converge to a specific value as more terms are added.

Geometric Progression

A geometric progression, or geometric sequence, is fundamental in understanding how series are built and develop. It comprises a set of numbers where each term is a product of the preceding term and a consistent ratio, not equal to zero. This ratio is known as the common ratio and is a crucial element because it dictates the growth or decay of the sequence.

In our example \( \sum_{i=1}^{6}\left(\frac{1}{3}\right)^{i+1} \), the series is geometric because every term is created by multiplying the previous term by \( \frac{1}{3} \). To grasp the whole sequence, you can write out the terms like this: \( \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, ... \), and so on up to the sixth term. Seeing the terms laid out helps picture the consistent decrease in size due to the ratio being less than one—indicative of a geometric sequence with a converging pattern.

Series Convergence

The concept of series convergence is a critical aspect of understanding how infinite geometric series behave. It answers whether a series approaches a finite value (converges) or grows without bounds (diverges) as more terms are added. For a geometric series to converge, its common ratio must be between -1 and 1, excluding these extremes. The sum formula \( S_n = \frac{a(1-r^n)}{1-r} \) works because it is based on the idea that the series is getting closer and closer to a certain value.

The convergence of the series can drastically affect the result. For example, if the common ratio were greater than 1 or less than -1, the terms would get progressively larger, and our sum would tend toward infinity, indicating divergence. However, in our exercise, since the common ratio \( \frac{1}{3} \) is between -1 and 1, the series converges, which is reflected in the finite sum of 0.13168.