Answer
(a) $ {2,\frac{{12}}{5},\frac{{62}}{{25}},\frac{{312}}{{125}}} $
${s_n} = \frac{1}{2}\left( {\frac{{{5^n} - 1}}{{{5^{n - 1}}}}} \right)$
Converges; $S=\frac{5}{2}$.
(b) $ {\frac{1}{4},\frac{3}{4},\frac{7}{4},\frac{{15}}{4}} $
${s_n} = \frac{{{2^n} - 1}}{4}$
Diverges
(c) ${\frac{1}{6},\frac{1}{4},\frac{3}{{10}},\frac{1}{3}} $
${s_n} = \frac{1}{2} - \frac{1}{{n + 2}}$
Converges; $S=\frac{1}{2}$.
Work Step by Step
(a) $2 + \dfrac{2}{5} + \dfrac{2}{{{5^2}}} + \cdot\cdot\cdot + \dfrac{2}{{{5^{k - 1}}}} + \cdot\cdot\cdot$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {2,2 + \dfrac{2}{5},2 + \dfrac{2}{5} + \dfrac{2}{{25}},2 + \dfrac{2}{5} + \dfrac{2}{{25}} + \dfrac{2}{{125}}} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {2,\dfrac{{12}}{5},\dfrac{{62}}{{25}},\dfrac{{312}}{{125}}} \right)$
(ii) A closed form for the $n$th partial sum:
The series can be written as $\mathop \sum \limits_{k = 1}^\infty 2{\left( {\dfrac{1}{5}} \right)^{k - 1}}$. So, the $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n 2{\left( {\dfrac{1}{5}} \right)^{k - 1}}$
Examining our results in part (i), we obtain the closed form for the $n$th partial sum:
${s_n} = \mathop \sum \limits_{k = 1}^n 2{\left( {\dfrac{1}{5}} \right)^{k - 1}} = \dfrac{1}{2}\left( {\dfrac{{{5^n} - 1}}{{{5^{n - 1}}}}} \right)$
(iii) Determine whether the series converges by calculating the limit of the $n$th partial sum.
We have ${s_n} = \mathop \sum \limits_{k = 1}^n 2{\left( {\dfrac{1}{5}} \right)^{k - 1}} = \dfrac{1}{2}\left( {\dfrac{{{5^n} - 1}}{{{5^{n - 1}}}}} \right)$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } \dfrac{{{5^n} - 1}}{{{5^{n - 1}}}} = \dfrac{1}{2}\mathop {\lim }\limits_{n \to \infty } 5\left( {\dfrac{{{5^n} - 1}}{{{5^n}}}} \right) = \dfrac{5}{2}\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{1}{{{5^n}}}} \right) = \dfrac{5}{2}$
So, the series converges and its sum is $\frac{5}{2}$.
(b) $\dfrac{1}{4} + \dfrac{2}{4} + \dfrac{{{2^2}}}{4} + \cdot\cdot\cdot + \dfrac{{{2^{k - 1}}}}{4} + \cdot\cdot\cdot$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4},\dfrac{1}{4} + \dfrac{2}{4},\dfrac{1}{4} + \dfrac{2}{4} + \dfrac{{{2^2}}}{4},\dfrac{1}{4} + \dfrac{2}{4} + \dfrac{{{2^2}}}{4} + \dfrac{{{2^3}}}{4}} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4},\dfrac{3}{4},\dfrac{7}{4},\dfrac{{15}}{4}} \right)$
(ii) A closed form for the $n$th partial sum:
The series can be written as $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{4}\left( {{2^{k - 1}}} \right)$. So, the $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{4}\left( {{2^{k - 1}}} \right)$
Examining our results in part (i), we obtain the closed form for the $n$th partial sum:
${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{4}\left( {{2^{k - 1}}} \right) = \dfrac{{{2^n} - 1}}{4}$
(iii) Determine whether the series converges by calculating the limit of the $n$th partial sum.
We have ${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{4}\left( {{2^{k - 1}}} \right) = \dfrac{{{2^n} - 1}}{4}$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{2^n} - 1}}{4} = \infty $
The limit does not exist. Therefore, the series diverges.
(c) $\dfrac{1}{{2\cdot 3}} + \dfrac{1}{{3\cdot 4}} + \dfrac{1}{{4\cdot 5}} + \cdot\cdot\cdot + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}} + \cdot\cdot\cdot$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{{2\cdot 3}},\dfrac{1}{{2\cdot 3}} + \dfrac{1}{{3\cdot 4}},\dfrac{1}{{2\cdot 3}} + \dfrac{1}{{3\cdot 4}} + \dfrac{1}{{4\cdot 5}},\dfrac{1}{{2\cdot 3}} + \dfrac{1}{{3\cdot 4}} + \dfrac{1}{{4\cdot 5}} + \dfrac{1}{{5\cdot 6}}} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{6},\dfrac{1}{4},\dfrac{3}{{10}},\dfrac{1}{3}} \right)$
(ii) A closed form for the $n$th partial sum:
The series can be written as $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$. So, the $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$
We can write $\dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{1}{{k + 1}} - \dfrac{1}{{k + 2}}$. So,
${s_n} = \mathop \sum \limits_{k = 1}^n \left( {\dfrac{1}{{k + 1}} - \dfrac{1}{{k + 2}}} \right)$
${s_n} = \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{5}} \right) + \cdot\cdot\cdot + \left( {\dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}} \right)$
${s_n} = \dfrac{1}{2} + \left( { - \dfrac{1}{3} + \dfrac{1}{3}} \right) + \left( { - \dfrac{1}{4} + \dfrac{1}{4}} \right) + \left( { - \dfrac{1}{5} + \dfrac{1}{5}} \right) + \cdot\cdot\cdot + \left( { - \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 1}}} \right) - \dfrac{1}{{n + 2}}$
${s_n} = \dfrac{1}{2} - \dfrac{1}{{n + 2}}$
Thus, the closed form for the $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{1}{2} - \dfrac{1}{{n + 2}}$
(iii) Determine whether the series converges:
We have ${s_n} = \mathop \sum \limits_{k = 1}^n \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{1}{2} - \dfrac{1}{{n + 2}}$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{2} - \dfrac{1}{{n + 2}}} \right) = \dfrac{1}{2}$
The limit exists. So, the series converges and its sum is $\frac{1}{2}$.
$\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{1}{2}$
