Question

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)

Accepted Answer

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 $$ -1 \u003c x \leq 1 . $$Determine the interval of convergence by applying the ratio test or by recalling the domain of the logarithm series. The series converges when $$ |x| \u003c 1 $$ 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 $$ \ln(\frac{1}{2}) .$$

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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Key Concepts

Maclaurin Series Expansion

Interval of Convergence

Evaluating Series at Specific Points