Evaluate the Summation sum from k=1 to infinity of -3(-1/3)^(k-1)
The problem presented is asking for the evaluation of an infinite series. Specifically, it wants you to determine the sum of the series that starts with "k=1" and goes to "infinity", where each term in the summation is given by the formula "-3(-1/3)^(k-1)". In other words, the question is asking you to find the total of an infinite number of terms, each term calculated by raising -1/3 to the power of (k-1) and then multiplying that result by -3. The series presented is a geometric series because each term is a constant times the previous term.
$\sum_{k = 1}^{\infty} - 3 \left(\left(\right. - \frac{1}{3} \left.\right)\right)^{k - 1}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, use the formula $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.
Step 2:
Identify the common ratio $r$ by using the formula $r = \frac{a_{k+1}}{a_k}$.
Step 2.1:
Insert $a_{k}$ and $a_{k+1}$ into the ratio formula: $r = \frac{-3(-\frac{1}{3})^{k}}{-3(-\frac{1}{3})^{k-1}}$.
Step 2.2:
Proceed to simplify the expression.
Step 2.2.1:
Eliminate the factor of $-3$: $r = \frac{\cancel{-3}(-\frac{1}{3})^{k}}{\cancel{-3}(-\frac{1}{3})^{k-1}}$.
Step 2.2.2:
Cancel out like terms in the numerator and denominator.
Step 2.2.2.1:
Extract $(-\frac{1}{3})^{k-1}$ from $(-\frac{1}{3})^{k}$: $r = \frac{(-\frac{1}{3})^{k-1}(-\frac{1}{3})^{1}}{(-\frac{1}{3})^{k-1}}$.
Step 2.2.2.2:
Simplify by canceling out common factors.
Step 2.2.3:
Combine like terms to find $r$: $r = (-\frac{1}{3})^{1}$.
Step 2.2.4:
Finalize the simplification to get $r = -\frac{1}{3}$.
Step 3:
Verify that the absolute value of $r$ is less than 1 to ensure the series converges.
Step 4:
Calculate the first term $a$ by substituting $k=1$ into the series term $-3(-\frac{1}{3})^{k-1}$.
Step 4.1:
Replace $k$ with $1$: $a = -3(-\frac{1}{3})^{1-1}$.
Step 4.2:
Simplify the expression to find $a$.
Step 4.2.1:
Subtract inside the exponent: $a = -3(-\frac{1}{3})^{0}$.
Step 4.2.2:
Apply the exponent rule to get $a = -3(1)$.
Step 4.2.3:
Finalize the calculation of the first term: $a = -3$.
Step 5:
Insert $r$ and $a$ into the summation formula: $S = \frac{-3}{1 - (-\frac{1}{3})}$.
Step 6:
Simplify the expression to find the sum.
Step 6.1:
Work on the denominator: $1 + \frac{1}{3}$.
Step 6.2:
Multiply by the reciprocal to simplify the fraction: $-3 \times \frac{3}{4}$.
Step 6.3:
Perform the multiplication to get the sum: $S = -\frac{9}{4}$.
Step 7:
Express the result in various forms: Exact form $-\frac{9}{4}$, Decimal form $-2.25$, Mixed number form $-2\frac{1}{4}$.
Knowledge Notes:
Infinite Geometric Series: An infinite geometric series is a sum of the form $a + ar + ar^2 + ar^3 + \ldots$, where $a$ is the first term and $r$ is the common ratio. The series converges to a finite sum if $|r| < 1$.
Sum Formula: The sum of a convergent infinite geometric series is given by $S = \frac{a}{1 - r}$.
Common Ratio: The common ratio $r$ between successive terms of a geometric series is found by $r = \frac{a_{k+1}}{a_k}$.
Convergence: For a geometric series to converge, the common ratio $r$ must satisfy $|r| < 1$.
Simplification: Simplifying expressions often involves canceling common factors, applying exponent rules, and combining like terms.
Exponent Rules: Important rules include $a^0 = 1$ for any nonzero $a$, and $(ab)^n = a^n b^n$.
Negative Exponents: A negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent, e.g., $a^{-n} = \frac{1}{a^n}$.
Absolute Value: The absolute value of a number $|a|$ is the non-negative value of $a$ without regard to its sign.
