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 $n$.

<Proof>\\
For a natural number $n$, let\\
$a _ { n } = \frac { 1 } { n + 1 } + \frac { 1 } { n + 2 } + \cdots + \frac { 1 } { 3 n + 1 }$.\\
We need 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 when $n = k$, $a _ { k } > 1$.
$$\begin{aligned}
& \text{When } n = k + 1 \\
& \begin{aligned}
a _ { k + 1 } & = \frac { 1 } { k + 2 } + \frac { 1 } { k + 3 } + \cdots + \frac { 1 } { 3 k + 4 } \\
& = a _ { k } + \left( \frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 3 } + \frac { 1 } { 3 k + 4 } \right)
\end{aligned}
\end{aligned}$$
On the other hand, $( 3 k + 2 ) ( 3 k + 4 )$ $\square$ (b) $( 3 k + 3 ) ^ { 2 }$, so
$$\frac { 1 } { 3 k + 2 } + \frac { 1 } { 3 k + 4 } > \text{(c)}$$
Since $a _ { k } > 1$,
$$a _ { k + 1 } > a _ { k } + \left( \frac { 1 } { 3 k + 3 } + \text{(c)} \right) \text{(a)} > 1$$
Therefore, by (1) and (2), $a _ { n } > 1$ for all natural numbers $n$.

What are the correct values for (a), (b), and (c) in the above proof? [3 points]

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
(a) & (b) & (c) \\
\hline
(1) $\frac { 1 } { k + 1 }$ & $>$ & $\frac { 2 } { 3 k + 3 }$ \\
\hline
(2) $\frac { 1 } { k + 1 }$ & $<$ & $\frac { 2 } { 3 k + 3 }$ \\
\hline
(3) $\frac { 1 } { k + 1 }$ & $<$ & $\frac { 4 } { 3 k + 3 }$ \\
\hline
(4) $\frac { 2 } { k + 1 }$ & $>$ & $\frac { 4 } { 3 k + 3 }$ \\
\hline
(5) $\frac { 2 } { k + 1 }$ & $<$ & $\frac { 1 } { k + 1 }$ \\
\hline
\end{tabular}
\end{center}