The mathematical expression in the image asks for the sum of the infinite series:

$\sum_{n=1}^\infty (2 - \sqrt{3})^n$

Solution:

This is a geometric series with the general form:

$\sum_{n=1}^\infty a^n = \frac{a}{1-a}, \quad \text{if } |a| < 1$

Here, $a = 2 - \sqrt{3}$.

Step 1: Check convergence

For the series to converge, we need $|2 - \sqrt{3}| < 1$.

Since $\sqrt{3} \approx 1.732$, we have:

$2 - \sqrt{3} \approx 0.268, \quad \text{and } |0.268| < 1.$

Thus, the series converges.

Step 2: Use the formula for the sum of a geometric series

The series starts from $n=1$, so the formula becomes:

$\sum_{n=1}^\infty (2 - \sqrt{3})^n = \frac{(2 - \sqrt{3})}{1 - (2 - \sqrt{3})}$

Step 3: Simplify the denominator

$1 - (2 - \sqrt{3}) = 1 - 2 + \sqrt{3} = \sqrt{3} - 1.$

Substitute back into the formula:

$\sum_{n=1}^\infty (2 - \sqrt{3})^n = \frac{2 - \sqrt{3}}{\sqrt{3} - 1}.$

Step 4: Rationalize the denominator

To simplify $\frac{2 - \sqrt{3}}{\sqrt{3} - 1}$, multiply numerator and denominator by $\sqrt{3} + 1$:

$\frac{2 - \sqrt{3}}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(2 - \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}.$

Simplify the denominator using $(\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2$.

Now simplify the numerator:

$(2 - \sqrt{3})(\sqrt{3} + 1) = 2\sqrt{3} + 2 - 3 - \sqrt{3} = (\sqrt{3} - 1).$

So the fraction becomes:

$\frac{\sqrt{3} - 1}{2}.$

Final Answer:

$\sum_{n=1}^\infty (2 - \sqrt{3})^n = \frac{\sqrt{3} - 1}{2}.$

Let me know if you'd like further details or clarifications!

Related Questions:

1. How can you verify if a series converges or diverges?
2. What is the general formula for a finite geometric series?
3. What happens if $|a| \geq 1$ in a geometric series?
4. Can you compute the sum of $\sum_{n=0}^\infty (2 - \sqrt{3})^n$?
5. How do you rationalize denominators with surds like $\sqrt{3} - 1$?

Tip:

Always check the convergence criteria ($|a| < 1$) before attempting to sum an infinite geometric series!