Evaluating an infinite series Write the Maclaurin series for f(x) = ln (1+x) and find the interval of convergence. Evaluate f(−1/2) to find the value of ∑ₖ₌₁∞ 1/(k 2ᵏ)

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Recall that the Maclaurin series for a function \( f(x) \) is its Taylor series expansion at \( x = 0 \), given by \( f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n \). For \( f(x) = \ln(1+x) \), start by finding the derivatives and evaluating them at 0 to identify the pattern of coefficients.

Use the known Maclaurin series expansion for \( \ln(1+x) \), which is \( \ln(1+x) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k} \). This series converges for \( -1 < x \leq 1 \).

Determine the interval of convergence by applying the ratio test or by recalling the domain of the logarithm series. The series converges when \( |x| < 1 \) and conditionally at \( x = 1 \), but diverges at \( x = -1 \).

To evaluate \( f(-\frac{1}{2}) = \ln(1 - \frac{1}{2}) = \ln(\frac{1}{2}) \), substitute \( x = -\frac{1}{2} \) into the series: \( \ln(1 - \frac{1}{2}) = \sum_{k=1}^\infty (-1)^{k+1} \frac{(-\frac{1}{2})^k}{k} \). Simplify the powers and signs to express the sum in terms of \( \sum_{k=1}^\infty \frac{1}{k 2^k} \).

Recognize that the series \( \sum_{k=1}^\infty \frac{1}{k 2^k} \) corresponds to the negative of the evaluated series (due to the alternating signs), so rearrange the expression to isolate \( \sum_{k=1}^\infty \frac{1}{k 2^k} \) and express it in terms of \( \ln(\frac{1}{2}) \).

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Maclaurin Series Expansion

The Maclaurin series is a Taylor series expansion of a function about x = 0. It expresses a function as an infinite sum of its derivatives at zero, multiplied by powers of x and divided by factorial terms. For ln(1+x), the series is derived by differentiating and evaluating at zero, providing a polynomial approximation valid near x = 0.

Interval of Convergence

The interval of convergence defines the range of x-values for which the infinite series converges to the function. It is found using tests like the ratio or root test. For ln(1+x), the series converges when -1 < x ≤ 1, ensuring the series sum accurately represents the function within this domain.

Evaluating Series at Specific Points

Substituting specific values of x into the Maclaurin series allows evaluation of infinite sums related to the function. For example, plugging x = -1/2 into the ln(1+x) series helps find the sum of ∑ₖ₌₁∞ 1/(k 2ᵏ). This technique connects function values to series sums, enabling exact or approximate calculations of infinite series.