Determine the interval of convergence for $\sum_{i=0}^\infty\frac{1+2^n}{1+3^n}x^n.$

Question

Determine the interval of convergence: $$\sum_{i=0}^\infty \frac{1+2^n}{1+3^n}x^n$$ Using Ratio test I found: $$-\frac{3}{2}<x<\frac{3}{2}$$ Now I don't know how to check for boundaries: $$For\ x = -\frac{3}{2}: \sum_{i=0}^\infty \frac{1+2^n}{1+3^n}(-\frac{3}{2})^n$$ Is this alternating series or alternating series is only when $$(-1)^n$$ ?

Answer 1

When $x=3/2$, we obtain the series $$\sum_{n=0}^{\infty}\frac{1+2^n}{1+3^n}\left(\frac 32\right)^n=\sum_{n=0}^{\infty}\frac{3^n+6^n}{2^n+6^n}$$ Note that $$\frac{3^n+6^n}{2^n+6^n}=\frac{6^n(1/2^n+1)}{6^n(1/3^n+1)}=\frac{1/2^n+1}{1/3^n+1}\to 1\,\,(n\to\infty) $$ so a necessary condition for convergence is not satisfied. When $x=-3/2$, the situation is similar: the limit of the summand is not $0$. In that case, the limit doesn't even exist as the term oscillates between $1$ and $-1$.