The following is a proof by mathematical induction that the inequality
$$\sum _ { i = 1 } ^ { 2 n + 1 } \frac { 1 } { n + i } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 } > 1$$
holds for all natural numbers $n$.

$\langle$Proof$\rangle$\\
For a natural number $n$, let\\
$a _ { n } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 }$.\\
It suffices to show that $a _ { n } > 1$.\\
(1) When $n = 1$, $a _ { 1 } = \frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } > 1$.\\
(2) Assume that $a _ { k } > 1$ when $n = k$.
$$\begin{aligned} & \text{When } n = k + 1, \\ & \begin{aligned} a _ { k + 1 } & = \frac { 1 } { k + 2 } + \frac { 1 } { k + 3 } + \cdots + \frac { 1 } { 3 k + 4 } \\ & = a _ { k } + \left( \frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 3 } + \frac { 1 } { 3 k + 4 } \right) \end{aligned} \end{aligned}$$
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On the other hand, $( 3 k + 2 ) ( 3 k + 4 )$ (나) $( 3 k + 3 ) ^ { 2 }$, so
$$\frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 4 } > \text{ (다) }$$
Since $a _ { k } > 1$,
$$a _ { k + 1 } > a _ { k } + \left( \frac { 1 } { 3 k + 3 } + \text{ (다) } \right) \text{ (가) } > 1$$
Therefore, by (1) and (2), $a _ { n } > 1$ for all natural numbers $n$.

What are the correct expressions for (가), (나), and (다) in the above proof? [3 points]\\
\begin{tabular}{ccc}
(가) & (나) & (다)\\
(1) $\frac { 1 } { k + 1 }$ & $<$ & $\frac { 2 } { 3 k + 3 }$\\
(2) $\frac { 1 } { k + 1 }$ & $>$ & $\frac { 2 } { 3 k + 3 }$\\
(3) $\frac { 1 } { k + 1 }$ & $<$ & $\frac { 2 } { 3 k + 4 }$\\
(4) $\frac { 2 } { k + 1 }$ & $>$ & $\frac { 2 } { 3 k + 4 }$\\
(5) $\frac { 2 } { k + 1 }$ & $<$ & $\frac { 2 } { 3 k + 3 }$
\end{tabular}