# 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}

Determine whether the given series is convergent or divergent. Explain your answer. If the series is convergent, find its sum.
$\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}$
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Given the series:
$\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}$
Rewriting the given series as sum of two series:
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{{3}^{n}}{{4}^{n}}+\frac{{2}^{n+1}}{{4}^{n}}\right)$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}}{{4}^{n}}+\sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n+1}}{{4}^{n}}$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{3}{4}{\right)}^{n}+\sum _{n=0}^{\mathrm{\infty }}\frac{2\cdot {2}^{n}}{{4}^{n}}$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{3}{4}{\right)}^{n}+\sum _{n=0}^{\mathrm{\infty }}2\left(\frac{2}{4}{\right)}^{n}$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{3}{4}{\right)}^{n}+\sum _{n=0}^{\mathrm{\infty }}2\left(\frac{1}{2}{\right)}^{n}$
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 $\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}$ 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:
$\sum _{n=0}^{\mathrm{\infty }}\left(\frac{3}{4}{\right)}^{n}=\frac{1}{1-\frac{3}{4}}$
$⇒\sum _{n=0}^{\mathrm{\infty }}\left(\frac{3}{4}{\right)}^{n}=\frac{1}{\frac{1}{4}}=4$
$\sum _{n=0}^{\mathrm{\infty }}2\left(\frac{1}{2}{\right)}^{n}=\frac{2}{\frac{1}{2}}=4$
The sum of the given series will be:
$\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=4+4$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{{3}^{n}+{2}^{n+1}}{{4}^{n}}=8$
The given series is converging.
The sum of series is 8

Jeffrey Jordon