Key Concepts and Formulas

• Taylor Series Expansion of $\ln(1-x)$: The Taylor series expansion of $\ln(1-x)$ around $x=0$ is given by $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots$, which converges for $|x| < 1$.
• Rationalizing the Denominator: To simplify fractions with irrational denominators, multiply the numerator and denominator by the conjugate of the denominator.
• Telescoping Series: A series where most of the terms cancel out, leading to a simpler sum. This concept is implicitly used when splitting the series.

Step-by-Step Solution

Step 1: Analyze and Simplify the Terms of the Series Let the given series be $S$. We examine the first few terms to find a pattern. $S = 1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\frac{49-20 \sqrt{6}}{180}+\ldots$ Let's simplify each term:

• Term 1: $1$
• Term 2: $\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}} = \frac{1}{2} \left(1 - \frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{1}{2} \left(1 - \sqrt{\frac{2}{3}}\right)$
• Term 3: $\frac{5-2 \sqrt{6}}{18}$. Notice that $(\sqrt{3}-\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}$. So, the term is $\frac{(\sqrt{3}-\sqrt{2})^2}{18} = \frac{1}{18} \left(\sqrt{3}-\sqrt{2}\right)^2$. We can rewrite this as $\frac{1}{18} \left(\sqrt{3}\left(1-\sqrt{\frac{2}{3}}\right)\right)^2 = \frac{1}{18} \cdot 3 \left(1-\sqrt{\frac{2}{3}}\right)^2 = \frac{1}{6} \left(1-\sqrt{\frac{2}{3}}\right)^2$.
• Term 4: $\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}$. Let's try to express the numerator in terms of $(\sqrt{3}-\sqrt{2})$. Consider $(\sqrt{3}-\sqrt{2})^3 = (\sqrt{3})^3 - 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 - (\sqrt{2})^3 = 3\sqrt{3} - 9\sqrt{2} + 6\sqrt{3} - 2\sqrt{2} = 9\sqrt{3} - 11\sqrt{2}$. So, the term is $\frac{(\sqrt{3}-\sqrt{2})^3}{36\sqrt{3}}$. We can rewrite this as $\frac{\left(\sqrt{3}(1-\sqrt{\frac{2}{3}})\right)^3}{36\sqrt{3}} = \frac{3\sqrt{3}\left(1-\sqrt{\frac{2}{3}}\right)^3}{36\sqrt{3}} = \frac{1}{12}\left(1-\sqrt{\frac{2}{3}}\right)^3$.
• Term 5: $\frac{49-20 \sqrt{6}}{180}$. Consider $(\sqrt{3}-\sqrt{2})^4 = ((\sqrt{3}-\sqrt{2})^2)^2 = (5-2\sqrt{6})^2 = 25 - 20\sqrt{6} + 24 = 49 - 20\sqrt{6}$. So, the term is $\frac{(\sqrt{3}-\sqrt{2})^4}{180} = \frac{\left(\sqrt{3}(1-\sqrt{\frac{2}{3}})\right)^4}{180} = \frac{9\left(1-\sqrt{\frac{2}{3}}\right)^4}{180} = \frac{1}{20}\left(1-\sqrt{\frac{2}{3}}\right)^4$.

The series can be rewritten as: $S = 1 + \frac{1}{2}\left(1-\sqrt{\frac{2}{3}}\right) + \frac{1}{6}\left(1-\sqrt{\frac{2}{3}}\right)^2 + \frac{1}{12}\left(1-\sqrt{\frac{2}{3}}\right)^3 + \frac{1}{20}\left(1-\sqrt{\frac{2}{3}}\right)^4 + \ldots$ The denominators are $1, 2, 6, 12, 20, \ldots$. We can see a pattern in the denominators of the terms involving $\left(1-\sqrt{\frac{2}{3}}\right)^n$: $2=1\cdot 2$, $6=2\cdot 3$, $12=3\cdot 4$, $20=4\cdot 5$. So, the $n$-th term (for $n \ge 1$) is $\frac{1}{n(n+1)} \left(1-\sqrt{\frac{2}{3}}\right)^n$.

Step 2: Introduce a Substitution to Simplify the Series Expression Let $x = 1 - \sqrt{\frac{2}{3}}$. The series becomes: $S = 1 + \frac{x}{1\cdot 2} + \frac{x^2}{2\cdot 3} + \frac{x^3}{3\cdot 4} + \frac{x^4}{4\cdot 5} + \ldots$ We can use the partial fraction decomposition $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. $S = 1 + \left(\frac{1}{1} - \frac{1}{2}\right)x + \left(\frac{1}{2} - \frac{1}{3}\right)x^2 + \left(\frac{1}{3} - \frac{1}{4}\right)x^3 + \left(\frac{1}{4} - \frac{1}{5}\right)x^4 + \ldots$

Step 3: Rearrange the Series and Relate to the Logarithmic Expansion We can regroup the terms: $S = 1 + \left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) - \left(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right)$ Recall the Taylor series for $\ln(1-x)$: $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ldots$ So, $x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots = -\ln(1-x)$.

The second part of the expression is $\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots$. This can be written as $\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) - x = -\ln(1-x) - x$.

Substituting these back into the expression for $S$: $S = 1 + (-\ln(1-x)) - (-\ln(1-x) - x)$ $S = 1 - \ln(1-x) + \ln(1-x) + x$ $S = 1 + x$ However, this is incorrect. Let's re-examine the regrouping.

Let's write the series as: $S = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = 1 + \sum_{n=1}^{\infty} \left(\frac{x^n}{n} - \frac{x^n}{n+1}\right)$ $S = 1 + \left(\frac{x}{1} - \frac{x}{2}\right) + \left(\frac{x^2}{2} - \frac{x^2}{3}\right) + \left(\frac{x^3}{3} - \frac{x^3}{4}\right) + \ldots$ $S = 1 + \left(\frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \ldots\right) - \left(\frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} + \ldots\right)$ The first parenthesis is $-\ln(1-x)$. The second parenthesis can be written as: $\frac{1}{x}\left(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right) = \frac{1}{x}\left(-\ln(1-x) - x\right) = -\frac{\ln(1-x)}{x} - 1$.

So, $S = 1 + (-\ln(1-x)) - \left(-\frac{\ln(1-x)}{x} - 1\right)$ $S = 1 - \ln(1-x) + \frac{\ln(1-x)}{x} + 1$ $S = 2 + \ln(1-x)\left(\frac{1}{x} - 1\right)$ $S = 2 + \left(\frac{1}{x} - 1\right) \ln(1-x)$

Step 4: Substitute Back the Original Value of $x$ and Simplify We have $x = 1 - \sqrt{\frac{2}{3}}$. $S = 2 + \left(\frac{1}{1-\sqrt{\frac{2}{3}}} - 1\right) \ln\left(1 - \left(1-\sqrt{\frac{2}{3}}\right)\right)$ $S = 2 + \left(\frac{1 - (1-\sqrt{\frac{2}{3}})}{1-\sqrt{\frac{2}{3}}}\right) \ln\left(\sqrt{\frac{2}{3}}\right)$ $S = 2 + \left(\frac{\sqrt{\frac{2}{3}}}{1-\sqrt{\frac{2}{3}}}\right) \ln\left(\sqrt{\frac{2}{3}}\right)$ To simplify the fraction $\frac{\sqrt{\frac{2}{3}}}{1-\sqrt{\frac{2}{3}}}$, we rationalize the denominator: $\frac{\sqrt{\frac{2}{3}}}{1-\sqrt{\frac{2}{3}}} = \frac{\sqrt{\frac{2}{3}}}{1-\sqrt{\frac{2}{3}}} \times \frac{1+\sqrt{\frac{2}{3}}}{1+\sqrt{\frac{2}{3}}} = \frac{\sqrt{\frac{2}{3}} + \frac{2}{3}}{1 - \frac{2}{3}} = \frac{\sqrt{\frac{2}{3}} + \frac{2}{3}}{\frac{1}{3}} = 3\sqrt{\frac{2}{3}} + 2$ Note that $3\sqrt{\frac{2}{3}} = 3 \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{3} \sqrt{3} \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{3}\sqrt{2} = \sqrt{6}$. So, the fraction simplifies to $\sqrt{6} + 2$.

Now, substitute this back into the expression for $S$: $S = 2 + (\sqrt{6} + 2) \ln\left(\sqrt{\frac{2}{3}}\right)$ We can rewrite $\sqrt{6}+2$ as $2\left(\frac{\sqrt{6}}{2} + 1\right) = 2\left(\sqrt{\frac{6}{4}} + 1\right) = 2\left(\sqrt{\frac{3}{2}} + 1\right)$. And $\ln\left(\sqrt{\frac{2}{3}}\right) = \ln\left(\left(\frac{2}{3}\right)^{1/2}\right) = \frac{1}{2} \ln\left(\frac{2}{3}\right)$.

So, $S = 2 + 2\left(\sqrt{\frac{3}{2}} + 1\right) \cdot \frac{1}{2} \ln\left(\frac{2}{3}\right)$ $S = 2 + \left(\sqrt{\frac{3}{2}} + 1\right) \ln\left(\frac{2}{3}\right)$

Step 5: Compare with the Given Expression and Find $a$ and $b$ We are given that $S = 2 + \left(\sqrt{\frac{b}{a}}+1\right) \log_e\left(\frac{a}{b}\right)$. Comparing our result $S = 2 + \left(\sqrt{\frac{3}{2}}+1\right) \ln\left(\frac{2}{3}\right)$ with the given expression, we have: $\sqrt{\frac{b}{a}} = \sqrt{\frac{3}{2}}$ and $\log_e\left(\frac{a}{b}\right) = \ln\left(\frac{2}{3}\right)$.

From $\sqrt{\frac{b}{a}} = \sqrt{\frac{3}{2}}$, we get $\frac{b}{a} = \frac{3}{2}$. From $\log_e\left(\frac{a}{b}\right) = \ln\left(\frac{2}{3}\right)$, we get $\frac{a}{b} = \frac{2}{3}$. These two equations are consistent.

We are given that $a$ and $b$ are integers with $\operatorname{gcd}(a, b)=1$. From $\frac{a}{b} = \frac{2}{3}$, we can set $a=2$ and $b=3$. Let's check the gcd: $\operatorname{gcd}(2, 3) = 1$. This satisfies the condition.

Step 6: Calculate the Final Value We need to find the value of $11a + 18b$. Substituting $a=2$ and $b=3$: $11a + 18b = 11(2) + 18(3) = 22 + 54 = 76$.

Common Mistakes & Tips

• Algebraic Errors: The simplification of the series terms and the subsequent algebraic manipulations are prone to errors. Double-check each step carefully.
• Misidentification of Series: Failing to recognize the pattern and relate it to the known Taylor series of $\ln(1-x)$ will make the problem very difficult.
• Incorrectly Applying Logarithm Properties: Ensure correct use of logarithm properties, especially when dealing with negative signs or reciprocal arguments. For example, $\ln(x^{-1}) = -\ln(x)$.

Summary The problem involves identifying a complex series and rewriting its terms to reveal a connection with the Taylor expansion of $\ln(1-x)$. By carefully simplifying the series and performing algebraic manipulations, we transformed the given infinite series into a form that could be directly compared with the provided expression $2+\left(\sqrt{\frac{b}{a}}+1\right) \log _e\left(\frac{a}{b}\right)$. This comparison allowed us to determine the integer values of $a$ and $b$ that satisfy the given conditions. Finally, we used these values to compute the required expression $11a + 18b$.

The final answer is $\boxed{76}$.