Answer
(a)${\frac{1}{4},\frac{5}{{16}},\frac{{21}}{{64}},\frac{{85}}{{256}}} $
${s_n} = \frac{1}{3}\left( {\frac{{{4^n} - 1}}{{{4^n}}}} \right)$
Converges; $S=\frac{1}{3}$.
(b) $ {1,5,21,85}$
${s_n} = \frac{1}{3}\left( {{4^n} - 1} \right)$
Diverges
(c) ${\frac{1}{{20}},\frac{1}{{12}},\frac{3}{{28}},\frac{1}{8}} $
${s_n} = \frac{1}{4} - \frac{1}{{n + 4}}$
Converges; $S=\frac{1}{4}$.
Work Step by Step
(a) $\mathop \sum \limits_{k = 1}^\infty {\left( {\dfrac{1}{4}} \right)^k}$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4},\dfrac{1}{4} + \dfrac{1}{{{4^2}}},\dfrac{1}{4} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{4^3}}},\dfrac{1}{4} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{4^3}}} + \dfrac{1}{{{4^4}}}} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4},\dfrac{5}{{16}},\dfrac{{21}}{{64}},\dfrac{{85}}{{256}}} \right)$
(ii) A closed form for the $n$th partial sum:
The $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n {\left( {\dfrac{1}{4}} \right)^k}$
Examining our results in part (i), we obtain the closed form for the $n$th partial sum:
${s_n} = \mathop \sum \limits_{k = 1}^n {\left( {\dfrac{1}{4}} \right)^k} = \dfrac{1}{3}\left( {\dfrac{{{4^n} - 1}}{{{4^n}}}} \right)$
(iii) Determine whether the series converges by calculating the limit of the $n$th partial sum.
We have ${s_n} = \dfrac{1}{3}\left( {\dfrac{{{4^n} - 1}}{{{4^n}}}} \right)$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \dfrac{1}{3}\mathop {\lim }\limits_{n \to \infty } \dfrac{{{4^n} - 1}}{{{4^n}}} = \dfrac{1}{3}\mathop {\lim }\limits_{n \to \infty } \left( {1 - \dfrac{1}{{{4^n}}}} \right) = \dfrac{1}{3}$
So, the series converges and its sum is $\frac{1}{3}$.
(b) $\mathop \sum \limits_{k = 1}^\infty {4^{k - 1}}$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {1,4,1 + 4 + 16,1 + 4 + 16 + 64} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {1,5,21,85} \right)$
(ii) A closed form for the $n$th partial sum:
The $n$th partial sum is
${s_n} = \mathop \sum \limits_{k = 1}^n {4^{k - 1}}$
Examining our results in part (i), we obtain the closed form for the $n$th partial sum:
${s_n} = \mathop \sum \limits_{k = 1}^n {4^{k - 1}} = \dfrac{1}{3}\left( {{4^n} - 1} \right)$
(iii) Determine whether the series converges by calculating the limit of the $n$th partial sum.
We have ${s_n} = \dfrac{1}{3}\left( {{4^n} - 1} \right)$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \dfrac{1}{3}\mathop {\lim }\limits_{n \to \infty } \left( {{4^n} - 1} \right) = \infty $
Since the limit does not exist, the series diverges.
(c) $\mathop \sum \limits_{k = 1}^\infty \left( {\dfrac{1}{{k + 3}} - \dfrac{1}{{k + 4}}} \right)$
(i) The first four partial sums:
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4} - \dfrac{1}{5},\left( {\dfrac{1}{4} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{6}} \right),\left( {\dfrac{1}{4} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{6} - \dfrac{1}{7}} \right),\left( {\dfrac{1}{4} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{6}} \right) + \left( {\dfrac{1}{6} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{7} - \dfrac{1}{8}} \right)} \right)$
(1) ${\ \ \ \ \ }$ $\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{4} - \dfrac{1}{5},\dfrac{1}{4} - \dfrac{1}{6},\dfrac{1}{4} - \dfrac{1}{7},\dfrac{1}{4} - \dfrac{1}{8}} \right)$
$\left( {{s_1},{s_2},{s_3},{s_4}} \right) = \left( {\dfrac{1}{{20}},\dfrac{1}{{12}},\dfrac{3}{{28}},\dfrac{1}{8}} \right)$
(ii) A closed form for the $n$th partial sum:
Examining our results, equation (1) in part (i), we obtain the closed form for the $n$th partial sum:
${s_n} = \dfrac{1}{4} - \dfrac{1}{{n + 4}}$
(iii) Determine whether the series converges:
We have ${s_n} = \dfrac{1}{4} - \dfrac{1}{{n + 4}}$.
Evaluate the limit:
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{4} - \dfrac{1}{{n + 4}}} \right) = \dfrac{1}{4}$
The limit exists. So, the series converges and its sum is $\frac{1}{4}$.
