Determine whether each geometric series converges or diverges. If it converges, find its sum.

Solution We compare each series to $\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}$.

(a) In this series $a=8$ and $r=\dfrac{2}{5}$. Since $\vert r\vert=\dfrac{2}{5}<1$, the series converges and $$\sum\limits_{k\,=\,1}^{\infty }8 \left( \frac{2}{5}\right) ^{k-1}=\frac{8}{1-\dfrac{2}{5}}=8 \left( \frac{5}{3}\right) =\frac{40}{3}$$

(b) Here, $a=1$ and $r=-\dfrac{5}{9}$. Since $\vert r\vert =\dfrac{5}{9}<1$, the series converges and $$\sum\limits_{k\,=\,1}^{\infty }\left( -\frac{5}{9}\right) ^{k-1}=\frac{1}{ 1-\left( -\dfrac{5}{9}\right) }=\frac{9}{14}$$

559

(c) Here, $a=3$ and $r=\dfrac{3}{2}$. Since $\vert r\vert =\dfrac{3}{2}>1$, the series $\sum\limits_{k\,=\,1}^{\infty}3 \left( \dfrac{3}{2}\right) ^{k-1}$ diverges.

(d) $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k}}$ is not in the form $\sum\limits_{k\,=\,1}^{\infty }ar^{k-1}$. To place it in this form, we proceed as follows: $$\begin{eqnarray*} &\sum\limits_{k\,=\,1}^{\infty }~\frac{1}{2^{k}}=\sum\limits_{k\,=\,1}^{ \infty }\left( \frac{1}{2}\right) ^{k} = \sum\limits_{k\,=\,1}^{\infty }\left[ \dfrac{1}{2}\cdot \left( \dfrac{1}{2} \right) ^{k-1}\right] \\[-13pt] &\hspace{2pc} \underset{\color{#0066A7}{{{\hbox{Write in the form}}}} {\color{#0066A7}{\hbox{\(\sum\limits_{k=1}^{\infty}ar^{k-1}\)}}}}{\color{#0066A7}{{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height8pt depth0pt}}\right.}}}} \end{eqnarray*}$$

So, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2^{k}}$ is a geometric series with $a=\dfrac{1}{2}$ and $r=\dfrac{1}{2}$. Since $\vert r\vert <1,$ the series converges, and its sum is $$\sum\limits_{k\,=\,1}^{\infty }\frac{1}{2^{k}}=\frac{\dfrac{1}{2}}{1-\dfrac{ 1}{2}}=1$$

which agrees with the sum we found earlier.

(e) $\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}$ starts at $0,$ so it is not in the form, $\sum\limits_{k\,=1}^{ \infty }ar^{k-1}$. To place it in this form, change the index to $l$, where $l=k+1.$ Then when $k=0,$ $l=1$ and $$\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1} = \sum\limits_{l=1}^{\infty }\left(\dfrac{1}{3}\right) ^{l-2}=\sum\limits_{l=1}^{\infty }\left(\dfrac{1}{3}\right) ^{-1}\left( \dfrac{1}{3}\right) ^{l-1}=\sum\limits_{l=1}^{\infty }3 \left( \dfrac{1}{3} \right)^{l-1}$$

That is, $\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}=\sum\limits_{l=1}^{\infty }3 \left( \dfrac{1}{3}\right) ^{l-1}$ is a geometric series with $a=3$ and $r=\dfrac{1}{3}.$ Since $\vert r\vert <1,$ the series converges, and its sum is $$\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}=\dfrac{3}{1-\dfrac{1}{3}}=\dfrac{9}{2}$$