Key Concepts and Formulas:

1. Partial Fraction Decomposition: This technique is used to express a rational function as a sum or difference of simpler rational functions. For a term of the form $\frac{1}{n(n+1)}$, the decomposition is $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. This is essential for simplifying terms in the series.
2. Maclaurin Series for $\ln(1+x)$: The Maclaurin series expansion for $\ln(1+x)$ is given by: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ This series converges for $-1 < x \le 1$. When $x=1$, we get the specific series for $\ln(2)$: $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
3. Properties of Logarithms: Key properties used for simplification include $m \ln a = \ln(a^m)$ and $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$. Also, $\ln e = 1$.

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Step-by-Step Solution:

1. Identify the General Term and Apply Partial Fraction Decomposition The given series is $S = \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \dots$ The general term of the series can be written as $T_n = (-1)^{n+1} \frac{1}{n(n+1)}$. We apply partial fraction decomposition to the magnitude of the general term: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ This decomposition is applied because it transforms each term into a difference of two simpler fractions, which is a standard technique for evaluating series, often leading to telescoping sums or recognizable patterns.

2. Rewrite the Series Using the Decomposition Substituting the partial fraction decomposition into the general term, we get: $T_n = (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$ The series $S$ can now be written as: $S = \sum_{n=1}^{\infty} (-1)^{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$ Expanding the first few terms to observe the pattern: For $n=1$: $(-1)^{2} \left( \frac{1}{1} - \frac{1}{2} \right) = 1 \left( 1 - \frac{1}{2} \right)$ For $n=2$: $(-1)^{3} \left( \frac{1}{2} - \frac{1}{3} \right) = -1 \left( \frac{1}{2} - \frac{1}{3} \right)$ For $n=3$: $(-1)^{4} \left( \frac{1}{3} - \frac{1}{4} \right) = 1 \left( \frac{1}{3} - \frac{1}{4} \right)$ For $n=4$: $(-1)^{5} \left( \frac{1}{4} - \frac{1}{5} \right) = -1 \left( \frac{1}{4} - \frac{1}{5} \right)$ So, the series becomes: $S = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{5} \right) + \dots$

3. Group Terms and Simplify Removing the parentheses and grouping like terms: $S = 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} + \frac{1}{5} + \frac{1}{5} - \dots$ Combining the identical terms: $S = 1 - 2\left(\frac{1}{2}\right) + 2\left(\frac{1}{3}\right) - 2\left(\frac{1}{4}\right) + 2\left(\frac{1}{5}\right) - \dots$ Factoring out $2$ from all terms except the first: $S = 1 - 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$ This simplification step is crucial as it transforms the series into a form that can be directly compared with known series expansions.

4. Relate to the Maclaurin Series of $\ln(1+x)$ We recall the Maclaurin series for $\ln(1+x)$ evaluated at $x=1$: $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots$ Now, consider the expression in the parentheses from Step 3: $\left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$. We can express this in terms of $\ln(2)$: $\ln(2) = 1 - \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$ Rearranging this equation, we find: $\left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right) = 1 - \ln(2)$ This step is vital because it allows us to substitute a known mathematical constant ($\ln(2)$) for the infinite series part.

5. Substitute Back and Final Calculation Substitute the result from Step 4 back into the expression for $S$ from Step 3: $S = 1 - 2 (1 - \ln(2))$ Distribute the $-2$: $S = 1 - 2 + 2 \ln(2)$ $S = -1 + 2 \ln(2)$ To match the given options, we use logarithm properties: $2 \ln(2) = \ln(2^2) = \ln(4)$ and $1 = \ln(e)$. $S = \ln(4) - \ln(e)$ $S = \ln\left(\frac{4}{e}\right)$ This result matches option (A).

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Common Mistakes & Tips:

• Sign Errors: Be meticulous with the alternating signs. Mistakes in distributing negative signs or combining terms can easily lead to an incorrect answer.
• Partial Fraction Decomposition Accuracy: Ensure the partial fraction decomposition is performed correctly. An error here propagates through the entire solution.
• Recognizing Series: Practice recognizing standard Maclaurin/Taylor series expansions. This problem specifically tests the recognition of the $\ln(1+x)$ series.
• Logarithm Properties: Be comfortable applying logarithm properties to simplify the final expression and match the answer format of the options.

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Summary:

The problem involves evaluating an infinite series by first decomposing its general term using partial fractions. This transforms the series into a form that, after careful algebraic manipulation, can be directly related to the Maclaurin series expansion of $\ln(1+x)$ evaluated at $x=1$. By substituting the value of $\ln(2)$ and applying logarithm properties, the sum of the series is found to be $\ln\left(\frac{4}{e}\right)$.

The final answer is $\boxed{\log {\,_e}\left( {{4 \over e}} \right)}$.