To solve the infinite series

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

let's break it down step by step:

Step 1: Simplify the given term

The general term of the series is: $a_n = \frac{(-1)^n \cdot 2^{n+1}}{3^n}.$

We can rewrite $2^{n+1}$ as $2 \cdot 2^n$: $a_n = \frac{(-1)^n \cdot 2 \cdot 2^n}{3^n}.$

Now simplify $\frac{2^n}{3^n}$ to $\left(\frac{2}{3}\right)^n$: $a_n = 2 \cdot (-1)^n \cdot \left(\frac{2}{3}\right)^n.$

So the series becomes: $S = \sum_{n=1}^\infty 2 \cdot (-1)^n \cdot \left(\frac{2}{3}\right)^n.$

Step 2: Factor out constants

Factor out the 2: $S = 2 \sum_{n=1}^\infty (-1)^n \cdot \left(\frac{2}{3}\right)^n.$

Let $x = \frac{2}{3}$. Then the series simplifies to: $S = 2 \sum_{n=1}^\infty (-1)^n \cdot x^n.$

Step 3: Recognize the alternating geometric series

The formula for an alternating geometric series is: $\sum_{n=1}^\infty (-1)^n \cdot x^n = -\frac{x}{1+x}, \quad \text{for } |x| < 1.$

Here, $x = \frac{2}{3}$, and $|x| = \frac{2}{3} < 1$. Thus, the formula applies. Substituting $x = \frac{2}{3}$: $\sum_{n=1}^\infty (-1)^n \cdot \left(\frac{2}{3}\right)^n = -\frac{\frac{2}{3}}{1 + \frac{2}{3}}.$

Step 4: Simplify the fraction

Simplify the denominator: $1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}.$

So the series becomes: $\sum_{n=1}^\infty (-1)^n \cdot \left(\frac{2}{3}\right)^n = -\frac{\frac{2}{3}}{\frac{5}{3}} = -\frac{2}{5}.$

Step 5: Multiply by the factor of 2

Finally, multiply by the factor of 2: $S = 2 \cdot \left(-\frac{2}{5}\right) = -\frac{4}{5}.$

Final Answer:

$\sum_{n=1}^\infty \frac{(-1)^n \cdot 2^{n+1}}{3^n} = -\frac{4}{5}.$

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

Follow-up Questions:

1. How does the alternating geometric series formula work for different values of $|x|$?
2. Why is the convergence of the series dependent on $|x| < 1$?
3. What happens if the series starts from $n=0$ instead of $n=1$?
4. How do you handle infinite series that are not geometric?
5. Can this method be applied to non-alternating series?

Tip:

When simplifying series, always check if it matches a standard form (e.g., geometric or telescoping) for quicker solutions!