The problem involves analyzing the convergence of the series:

$\sum_{n=1}^\infty (-1)^n \frac{2^n x^n}{13^n}$

The suggestion is to use the Generalized Limit Ratio Test. Let me analyze the series step by step.

Generalized Limit Ratio Test:

To apply the test, consider the general term of the series: $a_n = (-1)^n \frac{2^n x^n}{13^n}.$

We focus on the absolute value of $a_n$, ignoring the alternating sign: $|a_n| = \frac{2^n x^n}{13^n}.$

The ratio test requires evaluating: $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$

Substituting $a_n = \frac{2^n x^n}{13^n}$, we compute: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{\frac{2^{n+1} x^{n+1}}{13^{n+1}}}{\frac{2^n x^n}{13^n}} = \frac{2^{n+1} x^{n+1} \cdot 13^n}{13^{n+1} \cdot 2^n x^n}.$

Simplify the terms: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{2 \cdot x}{13}.$

Now take the limit as $n \to \infty$: $L = \frac{2 \cdot x}{13}.$

Convergence Criteria:

1. If $L < 1$, the series converges absolutely.
2. If $L > 1$, the series diverges.
3. If $L = 1$, the test is inconclusive.

From the computation: $L = \frac{2x}{13}.$

• For convergence: $\frac{2x}{13} < 1 \implies x < \frac{13}{2} = 6.5$.
• For divergence: $\frac{2x}{13} > 1 \implies x > 6.5$.

Conclusion:

• The series converges absolutely if $x < 6.5$.
• The series diverges if $x > 6.5$.
• If $x = 6.5$, the test is inconclusive, and further analysis is required (e.g., applying another test).

Would you like me to expand on any step or use another method to verify?

Related Questions:

1. What is the meaning of absolute and conditional convergence for alternating series?
2. How is the ratio test applied for alternating series in general?
3. Can you apply the root test to this series instead?
4. What happens when the ratio test is inconclusive (e.g., $L = 1$)?
5. What are the properties of power series convergence?

Tip:

When solving series, it's often helpful to analyze whether it's a power series (like this one, involving $x^n$), as convergence often depends on $x$.