# Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum. sum_{n=0}^inftyfrac{3^n+2^{n+1}}{4^n}

Series
Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum.
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$

2020-10-26

Given the series:
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$
Rewriting the given series as sum of two series:
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\left(\frac{3^n}{4^n}+\frac{2^{n+1}}{4^n}\right)$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty\frac{3^n}{4^n}+\sum_{n=0}^\infty\frac{2^{n+1}}{4^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty\frac{2\cdot2^n}{4^n}$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{2}{4})^n$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=\sum_{n=0}^\infty(\frac{3}{4})^n+\sum_{n=0}^\infty2(\frac{1}{2})^n$$
Hence we get sum of two geometric series:
$$r_1=\frac34$$
$$r_2=\frac12$$
Since both $$|r_1|<1,|r_2|<1$$
Both the individual geometric series converge.
The given series is sum of two converging, hence the given series $$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}$$ converges.
The sum of an infinite geometric series is given as:
$$S=\frac{\text{1st Term}}{1-r}$$
The sum of the series will be:
$$\sum_{n=0}^\infty(\frac34)^n=\frac{1}{1-\frac34}$$
$$\Rightarrow\sum_{n=0}^\infty(\frac34)^n=\frac{1}{\frac14}=4$$
$$\sum_{n=0}^\infty2(\frac12)^n=\frac{2}{\frac12}=4$$
The sum of the given series will be:
$$\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=4+4$$
$$\Rightarrow\sum_{n=0}^\infty\frac{3^n+2^{n+1}}{4^n}=8$$