Evaluate the Summation sum from j=1 to infinity of (-3/5)^j
The given problem asks to determine the value of an infinite series where the general term of the series is given by (-3/5)^j, with j starting from 1 and going to infinity. This requires you to calculate the sum of all terms of the form (-3/5)^j as j takes on each integer value from 1 to infinity. This is a typical question in calculus or mathematical analysis related to the concept of geometric series and convergence. To solve this problem, one would need to know how to evaluate the sum of an infinite geometric series and understand the conditions under which the series converges to a finite value.
$\sum_{j = 1}^{\infty} \left(\left(\right. - \frac{3}{5} \left.\right)\right)^{j}$
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:
Identify the common ratio $r$ by using the formula $r = \frac{a_{j+1}}{a_j}$ and perform the necessary calculations.
Step 2.1:
Insert the values for $a_j$ and $a_{j+1}$ into the ratio formula: $r = \frac{(-\frac{3}{5})^{j+1}}{(-\frac{3}{5})^j}$.
Step 2.2:
Eliminate the common terms in the ratio expression.
Step 2.2.1:
Extract $(-\frac{3}{5})^j$ from $(-\frac{3}{5})^{j+1}$: $r = \frac{(-\frac{3}{5})^j \cdot (-\frac{3}{5})}{(-\frac{3}{5})^j}$.
Step 2.2.2:
Simplify by removing common factors.
Step 2.2.2.1:
Multiply by the identity factor of $1$: $r = \frac{(-\frac{3}{5})^j \cdot (-\frac{3}{5})}{(-\frac{3}{5})^j \cdot 1}$.
Step 2.2.2.2:
Cancel out the common terms: $r = \frac{\cancel{(-\frac{3}{5})^j} \cdot (-\frac{3}{5})}{\cancel{(-\frac{3}{5})^j} \cdot 1}$.
Step 2.2.2.3:
Reformulate the ratio: $r = \frac{-\frac{3}{5}}{1}$.
Step 2.2.2.4:
Calculate the ratio by dividing: $r = -\frac{3}{5}$.
Step 3:
Confirm that the series is convergent since $|r| < 1$.
Step 4:
Calculate the first term $a$ by substituting $j = 1$ into the series expression.
Step 4.1:
Replace $j$ with $1$: $a = (-\frac{3}{5})^1$.
Step 4.2:
Simplify to find the first term: $a = -\frac{3}{5}$.
Step 5:
Insert the values of $a$ and $r$ into the sum formula: $S = \frac{-\frac{3}{5}}{1 - (-\frac{3}{5})}$.
Step 6:
Simplify the expression to find the sum.
Step 6.1:
Multiply the numerator by the reciprocal of the denominator: $S = -\frac{3}{5} \cdot \frac{1}{1 - (-\frac{3}{5})}$.
Step 6.2:
Simplify the denominator.
Step 6.2.1:
Apply the double negative: $S = -\frac{3}{5} \cdot \frac{1}{1 + \frac{3}{5}}$.
Step 6.2.2:
Express $1$ as a fraction with a common denominator: $S = -\frac{3}{5} \cdot \frac{1}{\frac{5}{5} + \frac{3}{5}}$.
Step 6.2.3:
Combine the fractions in the denominator: $S = -\frac{3}{5} \cdot \frac{1}{\frac{8}{5}}$.
Step 6.2.4:
Add the numbers in the denominator: $S = -\frac{3}{5} \cdot \frac{1}{\frac{8}{5}}$.
Step 6.3:
Multiply by the reciprocal of the denominator: $S = -\frac{3}{5} \cdot \frac{5}{8}$.
Step 6.4:
Simplify the multiplication: $S = -\frac{3}{5} \cdot \frac{5}{8}$.
Step 6.5:
Cancel the common factor of $5$.
Step 6.5.1:
Move the negative sign to the numerator: $S = \frac{-3}{5} \cdot \frac{5}{8}$.
Step 6.5.2:
Eliminate the common factor of $5$: $S = \frac{-3}{\cancel{5}} \cdot \frac{\cancel{5}}{8}$.
Step 6.5.3:
Recompose the expression: $S = -3 \cdot \frac{1}{8}$.
Step 6.6:
Combine the terms: $S = \frac{-3}{8}$.
Step 6.7:
Position the negative sign in front of the fraction: $S = -\frac{3}{8}$.
Step 7:
Present the result in various formats.
Exact Form: $S = -\frac{3}{8}$ Decimal Form: $S = -0.375$
Knowledge Notes:
To solve the problem of evaluating the summation of an infinite geometric series, it is important to understand the following concepts:
Infinite Geometric Series: An infinite geometric series is a series of the form $\sum_{j=1}^{\infty} ar^{j-1}$, where $a$ is the first term and $r$ is the common ratio between consecutive terms.
Sum Formula: The sum of an infinite geometric series can be found using the formula $S = \frac{a}{1 - r}$, provided that the absolute value of the common ratio $|r| < 1$ (which ensures convergence).
Common Ratio: The common ratio $r$ is the factor by which each term in the series is multiplied to get the next term. It can be found using the formula $r = \frac{a_{j+1}}{a_j}$.
Convergence Criteria: For an infinite geometric series to converge, the common ratio must satisfy $|r| < 1$. If $|r| \geq 1$, the series diverges.
Simplification: Algebraic simplification is used throughout the process to reduce expressions to their simplest form, including canceling common factors and combining like terms.
Negative Exponents: Negative exponents indicate the reciprocal of the base raised to the corresponding positive exponent, e.g., $a^{-n} = \frac{1}{a^n}$.
Fraction Arithmetic: Operations with fractions often require finding a common denominator, simplifying, and sometimes multiplying by the reciprocal.
Understanding these concepts is crucial for correctly applying the formula and simplifying the expressions to find the sum of the series.
