Evaluate the Summation sum from y=1 to infinity of 8(2/3)^y
The given problem is asking for an evaluation of an infinite series, which is a sum of terms following a specific pattern that theoretically extends indefinitely. The specific series provided has a starting index of y=1 and continues on to infinity. Each term in the series is defined as 8 * (2/3)^y, which means that every term is a product of 8 and (2/3) raised to the power of y, the current position in the sequence. The problem requires calculating the sum of all these terms as y increases from 1 to infinity. Since the series has a constant ratio between successive terms (namely, 2/3), this is an example of a geometric series, and the task is to determine whether this series converges to a finite sum and, if so, to find that sum.
$\sum_{y = 1}^{\infty} 8 \left(\left(\right. \frac{2}{3} \left.\right)\right)^{y}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, apply 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 $r = \frac{a_{y+1}}{a_y}$ and perform the necessary calculations.
Step 2.1:
Insert the values for $a_{y}$ and $a_{y+1}$ into the ratio formula: $r = \frac{8(\frac{2}{3})^{y+1}}{8(\frac{2}{3})^{y}}$.
Step 2.2:
Proceed to simplify the expression.
Step 2.2.1:
Eliminate the identical factor of $8$ from both numerator and denominator: $r = \frac{\cancel{8}(\frac{2}{3})^{y+1}}{\cancel{8}(\frac{2}{3})^{y}}$.
Step 2.2.1.1:
Rewrite the simplified expression: $r = \frac{(\frac{2}{3})^{y+1}}{(\frac{2}{3})^{y}}$.
Step 2.2.2:
Further simplify by canceling out common factors of $(\frac{2}{3})^{y}$.
Step 2.2.2.1:
Factor out $(\frac{2}{3})^{y}$ from $(\frac{2}{3})^{y+1}$: $r = \frac{(\frac{2}{3})^{y} \cdot \frac{2}{3}}{(\frac{2}{3})^{y}}$.
Step 2.2.2.2:
Cancel the common factors: $r = \frac{\cancel{(\frac{2}{3})^{y}} \cdot \frac{2}{3}}{\cancel{(\frac{2}{3})^{y}} \cdot 1}$.
Step 2.2.2.2.1:
Simplify the ratio to $r = \frac{2}{3}$.
Step 3:
Confirm that the series converges since $|r| < 1$.
Step 4:
Calculate the first term of the series by substituting $y = 1$.
Step 4.1:
Substitute $y = 1$ into the term $8(\frac{2}{3})^{y}$: $a = 8(\frac{2}{3})^{1}$.
Step 4.2:
Simplify the expression to find the first term.
Step 4.2.1:
Multiply $8$ by $\frac{2}{3}$ to get $a = \frac{8 \cdot 2}{3}$.
Step 4.2.2:
Perform the multiplication to obtain $a = \frac{16}{3}$.
Step 5:
Insert the values of $a$ and $r$ into the summation formula: $S = \frac{\frac{16}{3}}{1 - \frac{2}{3}}$.
Step 6:
Simplify the expression to find the sum of the series.
Step 6.1:
Multiply the numerator by the reciprocal of the denominator: $S = \frac{16}{3} \cdot \frac{1}{1 - \frac{2}{3}}$.
Step 6.2:
Simplify the denominator by finding a common denominator: $S = \frac{16}{3} \cdot \frac{1}{\frac{3}{3} - \frac{2}{3}}$.
Step 6.3:
Multiply the numerator by the reciprocal of the denominator: $S = \frac{16}{3} \cdot 3$.
Step 6.4:
Cancel the common factor of $3$ to get the final sum: $S = 16$.
Knowledge Notes:
Infinite Geometric Series: An infinite geometric series is a series of the form $a + ar + ar^2 + ar^3 + ...$ where $a$ is the first term and $r$ is the common ratio. The series converges to a finite sum if $|r| < 1$.
Sum of an Infinite Geometric Series: The sum of an infinite geometric series can be found using the formula $S = \frac{a}{1 - r}$, provided that $|r| < 1$.
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 can be found using the formula $r = \frac{a_{y+1}}{a_y}$.
Simplification: Simplifying expressions involves canceling out common factors, combining like terms, and performing arithmetic operations to reduce the expression to its simplest form.
Convergence of Series: A series converges if its sequence of partial sums approaches a finite limit. For a geometric series, this happens when the absolute value of the common ratio is less than one ($|r| < 1$).
