Evaluate the Summation sum from i=1 to infinity of (1/3)^i

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.