Determine whether each geometric series converges or diverges. If it converges, find its sum.
(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 \[ \sum\limits_{k\,=\,1}^{\infty }\left( -\frac{5}{9}\right) ^{k-1}=\frac{1}{ 1-\left( -\dfrac{5}{9}\right) }=\frac{9}{14} \]
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(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 \[ \sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{1}{3}\right) ^{k-1}=\dfrac{3}{1-\dfrac{1}{3}}=\dfrac{9}{2} \]