91Ó°ÊÓ

The Ratio Test states that if $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = L < 1$$, where $$a_n$$ is the nth term of the series, then the series converges. If $$L > 1$$ or doesn't exist, then the series diverges. To use the Ratio Test on this series, find the ratio of the $$(n+1)^{th}$$ term to the nth term and take the limit as n approaches infinity. $$(n+1)^{th}\text{ term:} \frac{2^{n+1}+3^{n+1}}{4^{n+1}}$$ $$n^{th}\text{ term:} \frac{2^n+3^n}{4^n}$$ Now, find the ratio of the terms and take the limit. $$\lim_{n\to\infty} \frac{\frac{2^{n+1}+3^{n+1}}{4^{n+1}}}{\frac{2^n+3^n}{4^n}}$$

To simplify the expression, multiply the top and bottom by $$\frac{1}{(2^n+3^n){4^n}}$$: $$\lim_{n\rightarrow\infty} \frac{(2^{n+1}+3^{n+1})(2^n+3^n)}{(4^{n+1})(4^n)}$$ Now, expand the numerator and simplify the denominator: $$\lim_{n\rightarrow\infty} \frac{2^{2n+1}+2^n3^n+2^n3^{n+1}+3^{2n+1}}{4^{2n+1}}$$ Next, let's divide each term of the numerator by the denominator: $$\lim_{n\rightarrow\infty} \frac{2^{2n+1}}{4^{2n+1}}+\frac{2^n3^n}{4^{2n+1}}+\frac{2^n3^{n+1}}{4^{2n+1}}+\frac{3^{2n+1}}{4^{2n+1}}$$ Now, simplify each term: $$\lim_{n\rightarrow\infty} \frac{1}{2}+\frac{3}{4^n}+\frac{3^2}{4^n}+\frac{3^n}{4^{n+1}}$$ Finally, take the limit as n approaches infinity: $$\frac{1}{2}+\lim_{n\rightarrow\infty} \frac{3}{4^n}+\lim_{n\rightarrow\infty} \frac{3^2}{4^n}+\lim_{n\rightarrow\infty} \frac{3^n}{4^{n+1}}$$ As n approaches infinity, any term containing $$\frac{3^n}{4^n}$$ or higher power in the denominator will approach to 0. $$L = \frac{1}{2}$$

Since the limit L is less than 1, according to the Ratio Test, the series converges. Therefore, the given series $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$ converges.