Evaluate the Summation sum from i=1 to infinity of (1/3)^i
The question asks to determine the total value obtained by adding up an infinite series of terms where each term is given by the expression (1/3)^i. The index 'i' starts at 1 and goes to infinity, implying that each successive term is a power of 1/3 and hence, the terms get progressively smaller. Specifically, this is a geometric series because each term is obtained by multiplying the previous term by a common ratio, which is 1/3 in this case. The problem is to evaluate or find the exact sum of this infinite geometric series.
$\sum_{i = 1}^{\infty} \left(\left(\right. \frac{1}{3} \left.\right)\right)^{i}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, use the sum formula $S = \frac{a}{1 - r}$, where $a$ is the initial term and $r$ is the common ratio.
Step 2:
Calculate the common ratio using the formula $r = \frac{a_{i + 1}}{a_{i}}$.
Step 2.1:
Insert the values for $a_{i}$ and $a_{i + 1}$ into the ratio formula: $r = \frac{(\frac{1}{3})^{i + 1}}{(\frac{1}{3})^{i}}$.
Step 2.2:
Eliminate the shared base in the ratio expression.
Step 2.2.1:
Extract $(\frac{1}{3})^{i}$ from $(\frac{1}{3})^{i + 1}$: $r = \frac{(\frac{1}{3})^{i} \times \frac{1}{3}}{(\frac{1}{3})^{i}}$.
Step 2.2.2:
Remove the common terms.
Step 2.2.2.1:
Multiply by unity: $r = \frac{(\frac{1}{3})^{i} \times \frac{1}{3}}{(\frac{1}{3})^{i} \times 1}$.
Step 2.2.2.2:
Cancel out like terms: $r = \frac{\cancel{(\frac{1}{3})^{i}} \times \frac{1}{3}}{\cancel{(\frac{1}{3})^{i}} \times 1}$.
Step 2.2.2.3:
Reformulate the ratio: $r = \frac{\frac{1}{3}}{1}$.
Step 2.2.2.4:
Divide $\frac{1}{3}$ by $1$: $r = \frac{1}{3}$.
Step 3:
Since $|r| < 1$, the series is convergent.
Step 4:
Identify the first term of the series by inserting the initial index value.
Step 4.1:
Replace $i$ with $1$ in $(\frac{1}{3})^{i}$: $a = (\frac{1}{3})^{1}$.
Step 4.2:
Simplify to find the first term: $a = \frac{1}{3}$.
Step 5:
Plug the values of $a$ and $r$ into the sum formula: $S = \frac{\frac{1}{3}}{1 - \frac{1}{3}}$.
Step 6:
Simplify the expression.
Step 6.1:
Multiply the numerator by the reciprocal of the denominator: $S = \frac{1}{3} \times \frac{1}{1 - \frac{1}{3}}$.
Step 6.2:
Simplify the denominator.
Step 6.2.1:
Express $1$ as a fraction with the same denominator: $S = \frac{1}{3} \times \frac{1}{\frac{3}{3} - \frac{1}{3}}$.
Step 6.2.2:
Combine into a single fraction: $S = \frac{1}{3} \times \frac{1}{\frac{2}{3}}$.
Step 6.2.3:
Subtract within the denominator: $S = \frac{1}{3} \times \frac{1}{\frac{2}{3}}$.
Step 6.3:
Multiply by the reciprocal of the denominator: $S = \frac{1}{3} \times (\frac{3}{2})$.
Step 6.4:
Perform the multiplication: $S = \frac{1}{3} \times \frac{3}{2}$.
Step 6.5:
Cancel out the common factor.
Step 6.5.1:
Eliminate the common factor: $S = \frac{1}{\cancel{3}} \times \frac{\cancel{3}}{2}$.
Step 6.5.2:
Finalize the simplified expression: $S = \frac{1}{2}$.
Step 7:
The sum can be presented in different formats.
Exact Form: $\frac{1}{2}$
Decimal Form: $0.5$
Knowledge Notes:
An infinite 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 ($r$).
The sum of an infinite geometric series converges to a finite value if the absolute value of the common ratio is less than one ($|r| < 1$).
The sum of a convergent geometric series is calculated using the formula $S = \frac{a}{1 - r}$, where $a$ is the first term of the series.
Simplifying expressions often involves finding common denominators, canceling out common factors, and performing arithmetic operations such as addition, subtraction, multiplication, and division.
The result of a series can be expressed in various forms, including exact fractions and decimal approximations.
