Evaluate the Summation sum from i=1 to infinity of 19*0.03^i
The problem provided is asking for the calculation of an infinite series sum. Specifically, it involves determining the sum of a geometric series that starts at i=1 and goes to infinity. Each term of the series is given by the formula 19 * (0.03)^i, where i represents each consecutive integer starting from 1. The question requires the application of formulas that pertain to the convergence of infinite geometric series and determining the sum if it converges. The sum can be evaluated if the common ratio of the series is between -1 and 1, which ensures the series converges to a finite value.
$\sum_{i = 1}^{\infty} 19 \cdot \left(0.03\right)^{i}$
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 first term and $r$ is the common ratio.
Step 2:
Calculate the common ratio ($r$) by using the ratio formula $r = \frac{a_{n+1}}{a_n}$.
Step 2.1:
Insert the terms $a_{n}$ and $a_{n+1}$ into the ratio formula: $r = \frac{19 \cdot (0.03)^{n+1}}{19 \cdot (0.03)^{n}}$.
Step 2.2:
Proceed to simplify the expression.
Step 2.2.1:
Eliminate the identical factor of $19$.
Step 2.2.1.1:
Remove the common factor: $r = \frac{\cancel{19} \cdot (0.03)^{n+1}}{\cancel{19} \cdot (0.03)^{n}}$.
Step 2.2.1.2:
Reformulate the expression: $r = \frac{(0.03)^{n+1}}{(0.03)^{n}}$.
Step 2.2.2:
Cancel out the common powers of $(0.03)^{n}$ and $(0.03)^{n+1}$.
Step 2.2.2.1:
Factor out $(0.03)^{n}$ from $(0.03)^{n+1}$: $r = \frac{(0.03)^{n} \cdot 0.03}{(0.03)^{n}}$.
Step 2.2.2.2:
Eliminate the common factors.
Step 2.2.2.2.1:
Multiply by the identity $1$: $r = \frac{(0.03)^{n} \cdot 0.03}{(0.03)^{n} \cdot 1}$.
Step 2.2.2.2.2:
Remove the common factor: $r = \frac{\cancel{(0.03)^{n}} \cdot 0.03}{\cancel{(0.03)^{n}} \cdot 1}$.
Step 2.2.2.2.3:
Rephrase the expression: $r = \frac{0.03}{1}$.
Step 2.2.2.2.4:
Divide $0.03$ by $1$: $r = 0.03$.
Step 3:
Given that $|r| < 1$, the series is convergent.
Step 4:
Identify the first term of the series by substituting the initial index value and simplifying.
Step 4.1:
Replace $i$ with $1$ in $19 \cdot (0.03)^{i}$: $a = 19 \cdot (0.03)^{1}$.
Step 4.2:
Simplify the calculation.
Step 4.2.1:
Compute the power: $a = 19 \cdot 0.03$.
Step 4.2.2:
Multiply $19$ by $0.03$: $a = 0.57$.
Step 5:
Insert the values of $r$ and $a$ into the summation formula: $S = \frac{0.57}{1 - 0.03}$.
Step 6:
Simplify the result.
Step 6.1:
Subtract $0.03$ from $1$: $S = \frac{0.57}{0.97}$.
Step 6.2:
Divide $0.57$ by $0.97$: $S \approx 0.58762886$.
Knowledge Notes:
To solve the given problem, we used the concept of an infinite geometric series. An infinite geometric series is a series of the form $a + ar + ar^2 + ar^3 + \ldots$ where $a$ is the first term and $r$ is the common ratio between successive terms. The series converges to a finite sum if the absolute value of the common ratio $|r|$ is less than $1$. The sum of such a convergent series is given by the formula $S = \frac{a}{1 - r}$.
The steps involved in solving the problem include:
Identifying the first term ($a$) and the common ratio ($r$) of the series.
Verifying that the series converges by ensuring that $|r| < 1$.
Applying the sum formula for an infinite geometric series to find the sum.
In this problem, the first term is found by substituting $i = 1$ into the given expression $19 \cdot (0.03)^i$, resulting in $a = 19 \cdot 0.03$. The common ratio is determined by dividing any term in the series by its preceding term, which simplifies to $r = 0.03$. Since the absolute value of $r$ is less than $1$, the series converges, and we can apply the sum formula to find the sum. After substituting the values of $a$ and $r$ into the formula and simplifying, we obtain the sum of the series.
