Evaluate the Summation sum from k=1 to infinity of 3(2/3)^k
The problem presented is asking for the evaluation of an infinite series, specifically the summation of a sequence where each term is given by the expression 3(2/3)^k, and the summation starts from k=1 and continues indefinitely to k=infinity. This is a request to calculate the limit of the partial sums of this sequence as the number of terms approaches infinity. The series looks like it could be geometric based on its form, where each term is a constant ratio times the previous term, and the task would be to determine whether this series converges to a finite value and to calculate that value if it does. If the series diverges, that would also be a pertinent conclusion.
$\sum_{k = 1}^{\infty} 3 \left(\left(\right. \frac{2}{3} \left.\right)\right)^{k}$
Answer
Solution:
Step:1
To determine the sum of an infinite geometric series, use the formula $S = \frac{a}{1 - r}$, where $a$ is the initial term and $r$ is the common ratio.
Step:2
Identify the common ratio by applying the formula $r = \frac{a_{k + 1}}{a_{k}}$ and carry out the necessary calculations.
Step:2.1
Insert $a_{k}$ and $a_{k + 1}$ into the formula for $r$: $r = \frac{3 \left(\frac{2}{3}\right)^{k + 1}}{3 \left(\frac{2}{3}\right)^{k}}$.
Step:2.2
Proceed to simplify the expression.
Step:2.2.1
Eliminate the common factor of $3$.
Step:2.2.1.1
Remove the common factor: $r = \frac{\cancel{3} \left(\frac{2}{3}\right)^{k + 1}}{\cancel{3} \left(\frac{2}{3}\right)^{k}}$.
Step:2.2.1.2
Reformulate the equation: $r = \frac{\left(\frac{2}{3}\right)^{k + 1}}{\left(\frac{2}{3}\right)^{k}}$.
Step:2.2.2
Cancel out the common powers of $\left(\frac{2}{3}\right)^{k}$.
Step:2.2.2.1
Extract $\left(\frac{2}{3}\right)^{k}$ from $\left(\frac{2}{3}\right)^{k + 1}$: $r = \frac{\left(\frac{2}{3}\right)^{k} \cdot \frac{2}{3}}{\left(\frac{2}{3}\right)^{k}}$.
Step:2.2.2.2
Eliminate the common terms.
Step:2.2.2.2.1
Multiply by 1: $r = \frac{\left(\frac{2}{3}\right)^{k} \cdot \frac{2}{3}}{\left(\frac{2}{3}\right)^{k} \cdot 1}$.
Step:2.2.2.2.2
Remove the common factor: $r = \frac{\cancel{\left(\frac{2}{3}\right)^{k}} \cdot \frac{2}{3}}{\cancel{\left(\frac{2}{3}\right)^{k}} \cdot 1}$.
Step:2.2.2.2.3
Rephrase the equation: $r = \frac{2}{3}$.
Step:3
Confirm the series converges since the absolute value of $r$ is less than 1.
Step:4
Calculate the first term of the series by substituting $k = 1$ and simplifying.
Step:4.1
Replace $k$ with $1$ in $3 \left(\frac{2}{3}\right)^{k}$: $a = 3 \left(\frac{2}{3}\right)^{1}$.
Step:4.2
Simplify the expression.
Step:4.2.1
Simplify: $a = 3 \cdot \frac{2}{3}$.
Step:4.2.2
Remove the common factor of $3$.
Step:4.2.2.1
Eliminate the common factor: $a = \cancel{3} \cdot \frac{2}{\cancel{3}}$.
Step:4.2.2.2
Restate the result: $a = 2$.
Step:5
Insert the values of $r$ and $a$ into the sum formula: $S = \frac{2}{1 - \frac{2}{3}}$.
Step:6
Complete the simplification.
Step:6.1
Simplify the denominator.
Step:6.1.1
Express 1 as a fraction with a common denominator: $S = \frac{2}{\frac{3}{3} - \frac{2}{3}}$.
Step:6.1.2
Combine the numerators over the common denominator: $S = \frac{2}{\frac{1}{3}}$.
Step:6.1.3
Subtract 2 from 3: $S = \frac{2}{\frac{1}{3}}$.
Step:6.2
Multiply the numerator by the reciprocal of the denominator: $S = 2 \cdot 3$.
Step:6.3
Calculate the product of 2 and 3: $S = 6$.
Knowledge Notes:
The problem involves evaluating the sum of an infinite geometric series. The relevant knowledge points for solving this problem include:
Infinite Geometric Series: An infinite geometric series has the form $a + ar + ar^2 + ar^3 + \ldots$, where $a$ is the first term and $r$ is the common ratio between consecutive terms. The series converges if $|r| < 1$.
Sum of an Infinite Geometric Series: The sum of a convergent infinite geometric series can be calculated using the formula $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.
Simplification: Simplifying expressions involves canceling common factors, combining like terms, and performing arithmetic operations to reduce the expression to its simplest form.
Convergence Criteria: For a geometric series to converge, the absolute value of the common ratio $r$ must be less than 1. If $|r| \geq 1$, the series diverges.
Common Ratio: The common ratio $r$ in a geometric series is the factor by which each term is multiplied to get the next term. It is found by dividing any term by its preceding term, i.e., $r = \frac{a_{k + 1}}{a_{k}}$.
Reciprocal Multiplication: When dividing by a fraction, you can multiply by its reciprocal. For example, $\frac{2}{\frac{1}{3}}$ is equivalent to $2 \cdot 3$.
By understanding these concepts, one can solve problems related to infinite geometric series and determine their convergence and sums.
