1. **State the problem:** We are given two infinite series: $$\sum_{n=1}^{\infty} \frac{(x+2)^n}{n^2} \quad \text{and} \quad \sum_{n=1}^{\infty} \frac{(x+1)^n}{3^n}$$ We want to analyze and understand these series. 2. **Recall important formulas and rules:** - The first series is a power series with terms involving $\frac{(x+2)^n}{n^2}$. - The second series is a geometric series with ratio $\frac{x+1}{3}$. For a geometric series $\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$ if $|r|<1$. 3. **Analyze the second series (geometric):** $$\sum_{n=1}^\infty \frac{(x+1)^n}{3^n} = \sum_{n=1}^\infty \left(\frac{x+1}{3}\right)^n$$ This is a geometric series with first term $a = \frac{x+1}{3}$ and common ratio $r = \frac{x+1}{3}$. The sum converges if $\left|\frac{x+1}{3}\right| < 1$, i.e., $-4 < x < 2$. The sum is: $$S = \frac{a}{1-r} = \frac{\frac{x+1}{3}}{1 - \frac{x+1}{3}} = \frac{\frac{x+1}{3}}{\frac{3-(x+1)}{3}} = \frac{x+1}{3 - (x+1)} = \frac{x+1}{2 - x}$$ 4. **Analyze the first series:** $$\sum_{n=1}^\infty \frac{(x+2)^n}{n^2}$$ This is a power series similar to the polylogarithm function $\mathrm{Li}_2(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$. Here, $z = x+2$. The series converges for $|x+2| \leq 1$ (radius of convergence 1). The sum can be expressed as: $$\mathrm{Li}_2(x+2)$$ which is a special function called the dilogarithm. 5. **Summary:** - The first series converges for $|x+2| \leq 1$ and equals $\mathrm{Li}_2(x+2)$. - The second series converges for $-4 < x < 2$ and equals $\frac{x+1}{2 - x}$. 6. **Final answers:** $$\sum_{n=1}^\infty \frac{(x+2)^n}{n^2} = \mathrm{Li}_2(x+2) \quad \text{for} \quad |x+2| \leq 1$$ $$\sum_{n=1}^\infty \frac{(x+1)^n}{3^n} = \frac{x+1}{2 - x} \quad \text{for} \quad -4 < x < 2$$ These expressions describe the sums of the infinite series given the domain restrictions for convergence.