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}

Question

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}}

Answer 1

Given the series:
$\sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}}$
Rewriting the given series as sum of two series:
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = \sum_{n = 0}^{\infty} (\frac{3^{n}}{4^{n}} + \frac{2^{n + 1}}{4^{n}})$
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = \sum_{n = 0}^{\infty} \frac{3^{n}}{4^{n}} + \sum_{n = 0}^{\infty} \frac{2^{n + 1}}{4^{n}}$
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = \sum_{n = 0}^{\infty} (\frac{3}{4})^{n} + \sum_{n = 0}^{\infty} \frac{2 \cdot 2^{n}}{4^{n}}$
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = \sum_{n = 0}^{\infty} (\frac{3}{4})^{n} + \sum_{n = 0}^{\infty} 2 (\frac{2}{4})^{n}$
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = \sum_{n = 0}^{\infty} (\frac{3}{4})^{n} + \sum_{n = 0}^{\infty} 2 (\frac{1}{2})^{n}$
Hence we get sum of two geometric series:
$r_{1} = \frac{3}{4}$
$r_{2} = \frac{1}{2}$
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:
$S = \frac{1st Term}{1 - r}$
The sum of the series will be:
$\sum_{n = 0}^{\infty} (\frac{3}{4})^{n} = \frac{1}{1 - \frac{3}{4}}$
$\Rightarrow \sum_{n = 0}^{\infty} (\frac{3}{4})^{n} = \frac{1}{\frac{1}{4}} = 4$
$\sum_{n = 0}^{\infty} 2 (\frac{1}{2})^{n} = \frac{2}{\frac{1}{2}} = 4$
The sum of the given series will be:
$\sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = 4 + 4$
$\Rightarrow \sum_{n = 0}^{\infty} \frac{3^{n} + 2^{n + 1}}{4^{n}} = 8$
Final Answer:
The given series is converging.
The sum of series is 8