Question

Determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. $$ \sum_{n=1}^{\infty} \frac{2^{n}}{4 n^{2}-1} $$

Short answer

The given series \( \sum_{n=1}^{\infty} \frac{2^{n}}{4 n^{2}-1} \) is divergent as per the ratio test.

Step-by-step solution

Apply the Ratio Test

The Ratio Test says if \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = L \) where \( a_n \) is the terms of the series, then our series will converge if L < 1 and diverge if L > 1. Here \( \frac{a_{n+1}}{a_{n}} = \frac{\frac{2^{n+1}}{4(n+1)^2 - 1}}{\frac{2^n}{4n^2 - 1}} = \frac{2^{n+1}(4n^2 - 1)}{2^n(4(n+1)^2 - 1)}= \frac{2 * 2^n * (4n^2-1)}{2^n * (4n^2 + 8n + 4 - 1)} \)

Simplify and find Limit

Upon simplifying we get \( \frac{2 * (4n^2-1)}{(4n^2 + 8n + 3)} \).\nNow we apply the limit \( \lim_{n \to \infty} \frac{2 * (4n^2-1)}{(4n^2 + 8n + 3)}\). For polynomial functions of the same degree, we can take the ratio of the highest degree terms. Hence, our limit simply comes to \( \frac{2 * 4}{4} = 2 \).

Conclusion

Since the limit L is 2 which is greater than 1, according to the Ratio Test, the series \( \sum_{n=1}^{\infty} \frac{2^{n}}{4 n^{2}-1}\) is divergent.