If $$1 + \frac{\sqrt{3} - \sqrt{2}}{2\sqrt{3}} + \frac{5 - 2\sqrt{6}}{18} + \frac{9\sqrt{3} - 11\sqrt{2}}{36\sqrt{3}} + \frac{49 - 20\sqrt{6}}{180} + \ldots$$ upto $$\infty = 2 + \left(\sqrt{\frac{b}{a}} + 1\right)\log_e\left(\frac{a}{b}\right)$$, where $$a$$ and $$b$$ are integers with $$\gcd(a, b) = 1$$, then $$11a + 18b$$ is equal to ______

Let's find the general term for the series (excluding the first '$$1$$' for now). We can break down the numerators and denominators to spot the pattern.

1. Identifying the pattern:

Let's check the powers of $$(\sqrt{3} - \sqrt{2})$$ for the numerators:

• $$N_1 = (\sqrt{3} - \sqrt{2})^1$$
• $$N_2 = (\sqrt{3} - \sqrt{2})^2 = 5 - 2\sqrt{6}$$
• $$N_3 = (\sqrt{3} - \sqrt{2})^3 = 9\sqrt{3} - 11\sqrt{2}$$
• $$N_4 = (\sqrt{3} - \sqrt{2})^4 = 49 - 20\sqrt{6}$$

Now let's structure the denominators to match the $$n$$-th index:

• $$D_1 = 2\sqrt{3} = 2 \times 1 \times (\sqrt{3})^1$$
• $$D_2 = 18 = 3 \times 2 \times (\sqrt{3})^2$$
• $$D_3 = 36\sqrt{3} = 4 \times 3 \times (\sqrt{3})^3$$
• $$D_4 = 180 = 5 \times 4 \times (\sqrt{3})^4$$

Combining these, the $$n$$-th fraction of the series is:

$$T_n = \frac{(\sqrt{3} - \sqrt{2})^n}{n(n+1)(\sqrt{3})^n} = \frac{1}{n(n+1)} \left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}} \right)^n$$

Let $$x = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}} = 1 - \sqrt{\frac{2}{3}}$$.

The entire sum can now be written as:

$$S = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}$$

2. Evaluating the sum:

Using partial fractions, we know $$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$. Split the sum into two parts:

$$S = 1 + \sum_{n=1}^{\infty} \left( \frac{x^n}{n} - \frac{x^n}{n+1} \right)$$

$$S = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1}$$

Using the standard logarithmic series $$\sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$$, we can evaluate both parts:

• First sum: $$-\ln(1-x)$$
• Second sum: $$\frac{1}{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = \frac{1}{x} \left( -\ln(1-x) - x \right)$$

Substitute these back into $$S$$:

$$S = 1 - \ln(1-x) - \left[ \frac{-\ln(1-x) - x}{x} \right]$$

$$S = 1 - \ln(1-x) + \frac{1}{x}\ln(1-x) + 1$$

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

3. Substituting $$x$$ back:

We know $$x = 1 - \sqrt{\frac{2}{3}}$$, which means $$1 - x = \sqrt{\frac{2}{3}}$$.

Let's find the factor outside the log:

$$\frac{1-x}{x} = \frac{\sqrt{\frac{2}{3}}}{1 - \sqrt{\frac{2}{3}}} = \frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}$$

Rationalize the denominator by multiplying by $$(\sqrt{3} + \sqrt{2})$$:

$$\frac{\sqrt{2}(\sqrt{3} + \sqrt{2})}{3 - 2} = \sqrt{6} + 2$$

Now plug everything into our expression for $$S$$:

$$S = 2 + (\sqrt{6} + 2) \ln\left( \sqrt{\frac{2}{3}} \right)$$

To match the form given in the question, pull the $$\frac{1}{2}$$ power from the log argument to the front:

$$S = 2 + \frac{\sqrt{6} + 2}{2} \ln\left( \frac{2}{3} \right)$$

$$S = 2 + \left( \frac{\sqrt{6}}{2} + 1 \right) \ln\left( \frac{2}{3} \right)$$

$$S = 2 + \left( \sqrt{\frac{6}{4}} + 1 \right) \ln\left( \frac{2}{3} \right)$$

$$S = 2 + \left( \sqrt{\frac{3}{2}} + 1 \right) \log_e\left( \frac{2}{3} \right)$$

4. Final Comparison:

Compare this to the given RHS: $$2 + \left(\sqrt{\frac{b}{a}} + 1\right)\log_e\left(\frac{a}{b}\right)$$

• $$\frac{a}{b} = \frac{2}{3} \implies a = 2, b = 3$$
• (Checking the condition: $$\text{gcd}(2, 3) = 1$$, which is valid).

We need to find $$11a + 18b$$:

$$11(2) + 18(3) = 22 + 54 = 76$$