1. **Stating the problem:** We need to study the nature of the series $$\sum_{n=0}^{-\infty} \frac{5^n}{(2n+5)!}$$ and $$\sum_{n=2}^{+\infty} \frac{(-1)^n}{7^{n^3} - 1}$$ Then, for the power series $$S(x) = \sum_{n \geq 1} \frac{(3n^2 + 3) x^n}{n^2 + n - 1},$$ find the radius of convergence $R$ and the domain of convergence $D$. 2. **First series:** The sum $$\sum_{n=0}^{-\infty} \frac{5^n}{(2n+5)!}$$ is not standard because the index goes from 0 to $-\infty$, which is not a usual direction for sums. Typically, sums go from a lower to a higher index. This likely is a typo or misinterpretation. Assuming it means $$\sum_{n=0}^{+\infty} \frac{5^n}{(2n+5)!},$$ we analyze this. - The general term is $$a_n = \frac{5^n}{(2n+5)!}$$. - Factorials grow faster than any exponential, so the terms tend to zero very fast. - By the ratio test: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{5^{n+1}}{(2(n+1)+5)!} \cdot \frac{(2n+5)!}{5^n} = 5 \lim_{n \to \infty} \frac{1}{(2n+7)(2n+6)} = 0 < 1,$$ so the series converges absolutely. 3. **Second series:** $$\sum_{n=2}^{+\infty} \frac{(-1)^n}{7^{n^3} - 1}$$ - The denominator grows extremely fast because of $7^{n^3}$. - The terms tend to zero very fast. - By comparison with $$\frac{1}{7^{n^3}}$$, which converges absolutely, this series converges absolutely. 4. **Power series $S(x)$:** $$S(x) = \sum_{n=1}^\infty \frac{(3n^2 + 3) x^n}{n^2 + n - 1}$$ - To find the radius of convergence $R$, use the root or ratio test. - Consider the general term: $$a_n = \frac{(3n^2 + 3)}{n^2 + n - 1} x^n$$ - For large $n$, $$\frac{3n^2 + 3}{n^2 + n - 1} \sim \frac{3n^2}{n^2} = 3$$ - So asymptotically, $$a_n \approx 3 x^n$$ - Using the root test: $$\lim_{n \to \infty} |a_n|^{1/n} = \lim_{n \to \infty} \left| \frac{(3n^2 + 3)}{n^2 + n - 1} x^n \right|^{1/n} = |x| \lim_{n \to \infty} \left( \frac{3n^2 + 3}{n^2 + n - 1} \right)^{1/n} = |x| \cdot 1 = |x|$$ - The radius of convergence is where this limit equals 1: $$|x| = 1 \implies R = 1$$ 5. **Domain of convergence $D$:** - The power series converges absolutely for $$|x| < 1$$. - At $$x = 1$$, the series becomes: $$\sum_{n=1}^\infty \frac{3n^2 + 3}{n^2 + n - 1}$$ which behaves like $$3$$ times a series that diverges (since terms do not tend to zero), so it diverges. - At $$x = -1$$, the series is: $$\sum_{n=1}^\infty \frac{(3n^2 + 3)(-1)^n}{n^2 + n - 1}$$ - The terms behave like $$3 (-1)^n$$ times a term tending to 1, so the terms do not tend to zero, so it diverges. - Therefore, the domain of convergence is: $$D = \{ x \in \mathbb{R} : |x| < 1 \}$$ **Final answers:** - The first series converges absolutely. - The second series converges absolutely. - The radius of convergence of $S(x)$ is $$R = 1$$. - The domain of convergence is $$D = \{ x : |x| < 1 \}$$.