Question

For what values of \(x\) does the geometric series $$f(x)=\sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}$$ converge? Solve \(f(x)=3\)

Short answer

Answer: There are no values of x for which the given geometric series converges and f(x) = 3.

Step-by-step solution

Determine the convergence range for the geometric series

To find the range of \(x\) values for which the series converges, we need to find the range of values for which the absolute value of the common ratio, \(\frac{1}{1+x}\), is less than 1: $$\left|\frac{1}{1+x}\right| < 1$$ We can solve this inequality for \(x\) to find the values that satisfy the convergence condition.

Solve the inequality for the convergence range

Let's solve the inequality to find the range of \(x\) values for which the series converges: $$\left|\frac{1}{1+x}\right| < 1$$ We can examine two separate cases: when \(\frac{1}{1+x} > 0\) and when \(\frac{1}{1+x} < 0\). Case 1: \(\frac{1}{1+x} > 0\). In this case, we can remove the absolute value and solve for \(x\): $$\frac{1}{1+x} < 1$$ $$1 < 1+x$$ $$x > 0$$ Case 2: \(\frac{1}{1+x} < 0\). In this case, we can remove the absolute value by multiplying both sides by \(-1\), and reversing the inequality sign: $$-\frac{1}{1+x} < 1$$ $$1+x > -1$$ $$x > -2$$ From the two cases, we can see that the geometric series converges for \(-2 < x < 0\).

Find the sum of the convergent series

Now that we know the convergence range, let's find the sum of the convergent series. The formula for the sum of a convergent geometric series is $$S = \frac{a_1}{1-r}$$ where \(a_1\) is the first term in the series, and \(r\) is the common ratio. In this case, \(a_1 = 1\), and \(r = \frac{1}{1+x}\). Substituting these values into the formula, we get $$f(x) = \frac{1}{1-\frac{1}{1+x}}$$

Solve the equation f(x) = 3

Now, we need to solve the equation \(f(x) = 3\) to find the specific values of \(x\) that satisfy the condition: $$\frac{1}{1-\frac{1}{1+x}} = 3$$ $$1-x = \frac{1}{3}(1+x)$$ $$3-3x = 1+x$$ $$2x = 2$$ $$x = 1$$ The solution of the equation \(f(x) = 3\) is \(x = 1\), but the geometric series converges only for \(-2 < x < 0\). Since \(x = 1\) does not fall within the convergence range, there are no values of \(x\) for which the given geometric series converges and \(f(x) = 3\).

Key concepts

Geometric Series

A geometric series is a sequence of terms where each term is a constant multiple, known as the common ratio, of the previous term. A typical form for a geometric series is: \[S = a + ar + ar^2 + ar^3 + \ldots\] where \(a\) is the first term and \(r\) is the common ratio. Understanding geometric series is crucial because their properties greatly simplify when we know the common ratio's value. They have interesting characteristics, particularly depending on whether the common ratio is greater or less than one in absolute value.
- If \(|r| < 1\), the geometric series converges to a finite sum.- If \(|r| \geq 1\), the series becomes divergent, which means it doesn't converge to a specific value.
Recognizing these properties allows us to predict the behavior of a series based on its common ratio and aids in determining the series' convergence.

Convergence Range

When studying geometric series, identifying the convergence range is essential. The convergence range specifies the values for which the series converges to a finite sum. To find this range, we impose a condition on the common ratio \(r\). Specifically, we require that: \[|r| < 1\] In the given geometric series \(f(x) = \sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}\), the common ratio is \(\frac{1}{1+x}\). Consequently, we must solve the inequality: \[\left|\frac{1}{1+x}\right| < 1\] The solution helps us determine the interval for the variable \(x\) within which the series converges. This involves solving two cases: whether the expression is positive or negative.
- For \(\frac{1}{1+x} > 0\), simplifying gives \(x > 0\). - For \(\frac{1}{1+x} < 0\), solving yields \(x > -2\).
Thus, we find that for \(-2 < x < 0\), the series converges, defining its convergence range.

Solving Inequalities

Solving inequalities is a crucial step when working with series where determining a range of values is necessary. In our task, we faced an inequality involving absolute values: \[\left|\frac{1}{1+x}\right| < 1\] To solve it, we considered both the subsection (\