Evaluate the Summation sum from p=0 to infinity of 5(3/4)^p

Solution:

Step:1

To determine the sum of an infinite geometric series, we use the formula $S = \frac{a}{1 - r}$, where $a$ is the initial term and $r$ is the common ratio.

Step:2

To identify the common ratio $r$, we employ the formula $r = \frac{a_{p + 1}}{a_{p}}$ and proceed to simplify it.

Step:2.1

Insert $a_{p}$ and $a_{p + 1}$ into the ratio formula: $r = \frac{5(\frac{3}{4})^{p + 1}}{5(\frac{3}{4})^{p}}$.

Step:2.2

Proceed to simplify the expression.

Step:2.2.1

Eliminate the common factor of $5$.

Step:2.2.1.1

Remove the common factor: $r = \frac{\cancel{5}(\frac{3}{4})^{p + 1}}{\cancel{5}(\frac{3}{4})^{p}}$.

Step:2.2.1.2

Reformulate the expression: $r = \frac{(\frac{3}{4})^{p + 1}}{(\frac{3}{4})^{p}}$.

Step:2.2.2

Extract and cancel the common powers of $(\frac{3}{4})^{p}$.

Step:2.2.2.1

Factor out $(\frac{3}{4})^{p}$ from $(\frac{3}{4})^{p + 1}$: $r = \frac{(\frac{3}{4})^{p} \cdot \frac{3}{4}}{(\frac{3}{4})^{p}}$.

Step:2.2.2.2

Eliminate the common factors.

Step:2.2.2.2.1

Multiply by $1$: $r = \frac{(\frac{3}{4})^{p} \cdot \frac{3}{4}}{(\frac{3}{4})^{p} \cdot 1}$.

Step:2.2.2.2.2

Remove the common factor: $r = \frac{\cancel{(\frac{3}{4})^{p}} \cdot \frac{3}{4}}{\cancel{(\frac{3}{4})^{p}} \cdot 1}$.

Step:2.2.2.2.3

Rephrase the expression: $r = \frac{\frac{3}{4}}{1}$.

Step:2.2.2.2.4

Divide $\frac{3}{4}$ by $1$: $r = \frac{3}{4}$.

Step:3

Given that $|r| < 1$, it is established that the series is convergent.

Step:4

To find the first term of the series, substitute the lower bound into the term and simplify.

Step:4.1

Replace $p$ with $0$ in the term $5(\frac{3}{4})^{p}$: $a = 5(\frac{3}{4})^{0}$.

Step:4.2

Simplify the term.

Step:4.2.1

Apply exponent rules to $\frac{3}{4}$: $a = 5 \cdot \frac{3^{0}}{4^{0}}$.

Step:4.2.2

Recognize that any number raised to the power of $0$ equals $1$: $a = 5 \cdot \frac{1}{4^{0}}$.

Step:4.2.3

Again, any number raised to the power of $0$ equals $1$: $a = 5 \cdot \frac{1}{1}$.

Step:4.2.4

Cancel out the common factor of $1$.

Step:4.2.4.1

Eliminate the common factor: $a = 5 \cdot \frac{\cancel{1}}{\cancel{1}}$.

Step:4.2.4.2

Restate the expression: $a = 5 \cdot 1$.

Step:4.2.5

Multiply $5$ by $1$: $a = 5$.

Step:5

Insert the values of $r$ and $a$ into the summation formula: $S = \frac{5}{1 - \frac{3}{4}}$.

Step:6

Carry out the simplification process.

Step:6.1

Simplify the denominator.

Step:6.1.1

Express $1$ as a fraction with the same denominator: $S = \frac{5}{\frac{4}{4} - \frac{3}{4}}$.

Step:6.1.2

Combine the numerators over the common denominator: $S = \frac{5}{\frac{4 - 3}{4}}$.

Step:6.1.3

Subtract $3$ from $4$: $S = \frac{5}{\frac{1}{4}}$.

Step:6.2

Multiply the numerator by the reciprocal of the denominator: $S = 5 \cdot 4$.

Step:6.3

Multiply $5$ by $4$: $S = 20$.

Knowledge Notes:

The problem involves evaluating the sum of an infinite geometric series, which 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. The formula for the sum of an infinite geometric series is $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio. This formula only applies when the absolute value of the common ratio is less than one ($|r| < 1$), ensuring that the series converges to a finite sum.

In this problem, the first term $a$ is given by $5$, and the common ratio $r$ is determined by the expression $(\frac{3}{4})^p$. By substituting $p=0$, we find the first term $a = 5$. To find the common ratio $r$, we look at the ratio of a term to its preceding term, which simplifies to $\frac{3}{4}$. Since the common ratio is less than one, the series converges, and we can apply the sum formula to find the sum of the series. After simplifying the expression, we find that the sum $S$ is $20$.