Evaluate the Summation sum from n=1 to infinity of 3(1/2)^(n-1)
The question asks you to determine the value of an infinite series, which is a sum of an infinite number of terms. Specifically, the series in question is a geometric series because each term is a constant multiple of the previous term. The series starts from n=1 and continues indefinitely (to infinity), and each term of the series is given by the expression 3(1/2)^(n-1). To evaluate the summation, you would need to apply the formula for the sum of an infinite geometric series, which requires that the common ratio (in this case, 1/2) be between -1 and 1. The question is asking for the sum of this particular geometric series.
$\sum_{n = 1}^{\infty} 3 \left(\left(\right. \frac{1}{2} \left.\right)\right)^{n - 1}$
Answer
Solution:
Step:1
To determine the sum of an infinite geometric series, use the formula $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.
Step:2
Identify the common ratio $r$ by using the formula $r = \frac{a_{n+1}}{a_n}$.
Step:2.1
Insert $a_{n}$ and $a_{n+1}$ into the ratio formula: $r = \frac{3(\frac{1}{2})^{n}}{3(\frac{1}{2})^{n-1}}$.
Step:2.2
Proceed to simplify the ratio.
Step:2.2.1
Remove the constant $3$ from both numerator and denominator: $r = \frac{\cancel{3}(\frac{1}{2})^{n}}{\cancel{3}(\frac{1}{2})^{n-1}}$.
Step:2.2.2
Eliminate the common base of the exponentials by factoring: $r = \frac{(\frac{1}{2})^{n-1}(\frac{1}{2})^{1}}{(\frac{1}{2})^{n-1}}$.
Step:2.2.3
Cancel out the common terms: $r = \frac{(\frac{1}{2})^{1}}{1}$.
Step:2.2.4
Conclude that $r = \frac{1}{2}$.
Step:3
Confirm that the series converges by checking that the absolute value of $r$ is less than $1$.
Step:4
Calculate the first term $a$ by setting $n=1$ in the general term: $a = 3(\frac{1}{2})^{1-1}$.
Step:4.1
Simplify the exponent: $a = 3(\frac{1}{2})^{0}$.
Step:4.2
Recognize that any number to the power of $0$ equals $1$: $a = 3 \cdot 1$.
Step:4.3
Therefore, $a = 3$.
Step:5
Plug the values of $a$ and $r$ into the sum formula: $S = \frac{3}{1 - \frac{1}{2}}$.
Step:6
Simplify the expression to find the sum.
Step:6.1
Find a common denominator for the terms in the denominator: $S = \frac{3}{\frac{2}{2} - \frac{1}{2}}$.
Step:6.2
Combine the fractions in the denominator: $S = \frac{3}{\frac{1}{2}}$.
Step:6.3
Multiply by the reciprocal of the denominator: $S = 3 \cdot 2$.
Step:6.4
Finally, $S = 6$.
Knowledge Notes:
Infinite Geometric Series: An infinite geometric series is a series of the form $a + ar + ar^2 + ar^3 + \ldots$ where $a$ is the first term and $r$ is the common ratio. The series converges if $|r| < 1$ and its sum can be found using the formula $S = \frac{a}{1 - r}$.
Common Ratio: In a geometric series, the common ratio $r$ is the factor by which each term is multiplied to get the next term. It is found by dividing any term by the preceding term: $r = \frac{a_{n+1}}{a_n}$.
Convergence of Series: A series converges if the sum of its terms approaches a finite limit as the number of terms increases. For a geometric series, this happens when the absolute value of the common ratio is less than one, i.e., $|r| < 1$.
Simplification of Exponents: When simplifying expressions with exponents, remember that any number raised to the power of zero is one, and that you can cancel out common bases with exponents by subtracting the exponents.
Reciprocal Multiplication: To divide by a fraction, you can multiply by its reciprocal. For example, $\frac{a}{\frac{b}{c}} = a \cdot \frac{c}{b}$.
By applying these principles, we can evaluate the sum of the given infinite geometric series.
