Geometric series

Examples

Determine if the geometric series converges or diverges. If it converges, find the value:

$\sum_{n = 1}^{\infty} \frac{3^{n - 1}}{5^{n + 1}}$

\begin{aligned} \sum_{n = 1}^{\infty} \frac{3^{n - 1}}{5^{n + 1}} & = \sum_{n = 1}^{\infty} (\frac{1}{5^{2}}) (\frac{3^{n - 1}}{5^{n - 1}}) \\ = \sum_{n = 1}^{\infty} (\frac{1}{25}) {(\frac{3}{5})}^{n - 1} \end{aligned}

Since $| \frac{3}{5} | < 1$ , the geometric series is convergent.

\begin{aligned} = \frac{\frac{1}{25}}{1 - \frac{3}{5}} \\ = \frac{1}{10} \end{aligned}

$\sum_{n = 3}^{\infty} \frac{e^{n}}{π^{n - 2}}$

\begin{aligned} \sum_{n = 3}^{\infty} \frac{e^{n}}{π^{n - 2}} & = \sum_{n = 3}^{\infty} (\frac{e}{π}) (\frac{e^{n - 1}}{π^{n - 1}}) \\ = \sum_{n = 3}^{\infty} (\frac{e}{π}) {(\frac{e}{π})}^{n - 1} \\ = \sum_{n = 1}^{n} (\frac{e}{π}) {(\frac{e}{π})}^{n - 1} - \frac{e}{π} - {(\frac{e}{π})}^{2} \end{aligned}

Since $| \frac{e}{π} | < 1$ , the geometric series is convergent.

\begin{aligned} = \frac{\frac{e}{π}}{1 - \frac{e}{π}} - \frac{e}{π} - {(\frac{e}{π})}^{2} \\ = \frac{e}{π - e} - \frac{e}{π} - {(\frac{e}{π})}^{2} \end{aligned}

$\sum_{n = 1}^{\infty} 2^{2 n} 3^{1 - n}$

\begin{aligned} \sum_{n = 1}^{\infty} 2^{2 n} 3^{1 - n} & = \sum_{n = 1}^{\infty} 4 \cdot 4^{n - 1} \cdot {\frac{1}{3}}^{n - 1} \\ = \sum_{n = 1}^{\infty} 4 \cdot {\frac{4}{3}}^{n - 1} \end{aligned}

Since $| \frac{4}{3} | \geq 1$ , the geometric series is divergent.

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