Evaluate the Summation sum from k=1 to infinity of (-1/2)^k
The problem provided involves calculating an infinite series. Specifically, it asks for the evaluation of the summation of a sequence where each term is given by (-1/2)^k as k increases from 1 to infinity. The series is alternating, as the term (-1/2) raised to any integer power will alternate between positive and negative values. The goal is to determine the sum of this series if it converges toward a particular number as more and more terms are added together indefinitely.
$\sum_{k = 1}^{\infty} \left(\left(\right. - \frac{1}{2} \left.\right)\right)^{k}$
Answer
Solution:
Step 1:
To determine the sum of an infinite geometric series, use the formula $S = \frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio.
Step 2:
Identify the common ratio by applying the formula $r = \frac{a_{k+1}}{a_k}$.
Step 2.1:
Insert $a_k$ and $a_{k+1}$ into the ratio formula: $r = \frac{(-\frac{1}{2})^{k+1}}{(-\frac{1}{2})^k}$.
Step 2.2:
Eliminate the shared base of $(-\frac{1}{2})^{k+1}$ and $(-\frac{1}{2})^k$.
Step 2.2.1:
Extract $(-\frac{1}{2})^k$ from $(-\frac{1}{2})^{k+1}$: $r = \frac{(-\frac{1}{2})^k \cdot (-\frac{1}{2})}{(-\frac{1}{2})^k}$.
Step 2.2.2:
Remove the common terms.
Step 2.2.2.1:
Introduce a multiplicative identity: $r = \frac{(-\frac{1}{2})^k \cdot (-\frac{1}{2})}{(-\frac{1}{2})^k \cdot 1}$.
Step 2.2.2.2:
Discard the shared factor: $r = \frac{\cancel{(-\frac{1}{2})^k} \cdot (-\frac{1}{2})}{\cancel{(-\frac{1}{2})^k} \cdot 1}$.
Step 2.2.2.3:
Reformulate the ratio: $r = \frac{-\frac{1}{2}}{1}$.
Step 2.2.2.4:
Compute the division: $r = -\frac{1}{2}$.
Step 3:
Confirm the series convergence by ensuring $|r| < 1$.
Step 4:
Compute the first term by substituting the initial index into the sequence.
Step 4.1:
Replace $k$ with $1$ in $(-\frac{1}{2})^k$: $a = (-\frac{1}{2})^1$.
Step 4.2:
Simplify to find the first term: $a = -\frac{1}{2}$.
Step 5:
Insert the first term and ratio into the summation formula: $S = \frac{-\frac{1}{2}}{1 - (-\frac{1}{2})}$.
Step 6:
Perform simplification of the expression.
Step 6.1:
Multiply the numerator by the reciprocal of the denominator: $S = -\frac{1}{2} \cdot \frac{1}{1 - (-\frac{1}{2})}$.
Step 6.2:
Resolve the denominator.
Step 6.2.1:
Double negation multiplication: $S = -\frac{1}{2} \cdot \frac{1}{1 + \frac{1}{2}}$.
Step 6.2.2:
Express 1 as a fraction with a common denominator: $S = -\frac{1}{2} \cdot \frac{1}{\frac{2}{2} + \frac{1}{2}}$.
Step 6.2.3:
Combine fractions over the common denominator: $S = -\frac{1}{2} \cdot \frac{1}{\frac{3}{2}}$.
Step 6.2.4:
Add the numerators: $S = -\frac{1}{2} \cdot \frac{2}{3}$.
Step 6.3:
Multiply the numerator by the reciprocal of the denominator: $S = -\frac{1}{2} \cdot \frac{2}{3}$.
Step 6.4:
Perform the multiplication: $S = -\frac{1}{2} \cdot \frac{2}{3}$.
Step 6.5:
Simplify by canceling out common factors.
Step 6.5.1:
Incorporate the negative sign into the numerator: $S = \frac{-1}{2} \cdot \frac{2}{3}$.
Step 6.5.2:
Cancel out the common factor of 2: $S = \frac{-1}{\cancel{2}} \cdot \frac{\cancel{2}}{3}$.
Step 6.5.3:
Finalize the expression: $S = -\frac{1}{3}$.
Step 7:
Present the result in various formats.
- Exact Form: $S = -\frac{1}{3}$
- Decimal Form: $S \approx -0.333$
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 sum of such a series can be found if the absolute value of the common ratio is less than one, indicating the series converges.
The formula for the sum of an infinite geometric series is $S = \frac{a}{1 - r}$, where:
$S$ is the sum of the series.
$a$ is the first term of the series.
$r$ is the common ratio between consecutive terms.
To find the common ratio $r$, we use the formula $r = \frac{a_{k+1}}{a_k}$, which involves dividing any term in the series by its preceding term.
In this problem, the series is given by the summation of $(-\frac{1}{2})^k$ from $k=1$ to infinity. The first term $a$ is obtained by substituting $k=1$ into the sequence, resulting in $a = -\frac{1}{2}$. The common ratio $r$ is determined by the expression $(-\frac{1}{2})^{k+1} / (-\frac{1}{2})^k$, which simplifies to $r = -\frac{1}{2}$.
Once $a$ and $r$ are known, they are substituted into the sum formula to find the sum of the series. The final result is simplified through algebraic manipulation, including multiplying by the reciprocal and canceling common factors, to arrive at the exact sum, which can also be expressed as a decimal.
