, 2022

We note that $$\begin{align*} t_1 & = \dfrac{1}{1} - \dfrac{1}{3} = \dfrac{2}{3} \approx 0.67\\ t_1 + t_2 & = \left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) = \dfrac{2}{3} + \dfrac{1}{4} = \dfrac{11}{12} \approx 0.92\\ t_1 + t_2 + t_3 & = \left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right)\\ & = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{5} \\ & = \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{4} - \dfrac{1}{5} = 1.05\\ t_1 + t_2 + t_3 + t_4 & = \left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \left(\dfrac{1}{4} - \dfrac{1}{6}\right)\\ & = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{5} - \dfrac{1}{6}\\ & = \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{5} - \dfrac{1}{6} \approx 1.13\end{align*}$$ This means that the sum of the first
$k$ terms is less than 1.499 for
$k = 1,2,3,4$.

When $k > 4$, we can extend the
pattern that we saw for $k = 3$ and
$k = 4$ to note that $$\begin{align*} t_1 + t_2 + t_3 + \ldots + t_{k-1} + t_k & = \left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) + \cdots \\ & + \left(\dfrac{1}{k-1} - \dfrac{1}{k+1}\right) + \left(\dfrac{1}{k} - \dfrac{1}{k+2}\right) \\ & = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{k-1} + \dfrac{1}{k} - \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{5} - \cdots - \dfrac{1}{k+1} - \dfrac{1}{k+2}\\ & = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{3} + \cdots + \dfrac{1}{k-1} - \dfrac{1}{k-1} + \dfrac{1}{k} - \dfrac{1}{k} - \dfrac{1}{k+1} - \dfrac{1}{k+2}\\ & = \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{k+1} - \dfrac{1}{k+2} \\ & = 1.500 - \dfrac{1}{k+1} - \dfrac{1}{k+2} \end{align*}$$
This means that the sum of the first $k$ terms is less than 1.499 exactly when
$\dfrac{1}{k+1} + \dfrac{1}{k+2}$ is
greater than 0.001.

As $k$ increases from 4, each of
$\dfrac{1}{k+1}$ and $\dfrac{1}{k+2}$ decreases, which means
that their sum decreases as well.

When $k = 1998$, $\$\dfrac{1}{k+1} + \dfrac{1}{k+2} = \dfrac{1}{1999} + \dfrac{1}{2000} > \dfrac{1}{2000} + \dfrac{1}{2000} = \dfrac{1}{1000}$ = 0.001$.

When $k = 1999$, $\$\dfrac{1}{k+1} + \dfrac{1}{k+2} = \dfrac{1}{2000} + \dfrac{1}{2001} < \dfrac{1}{2000} + \dfrac{1}{2000} = \dfrac{1}{1000}$ = 0.001$.

This means that $\$\dfrac{1}{k+1} + \dfrac{1}{k+2}$ is greater than 0.001 exactly when k $\leq$ 1998$ and is less than 0.001 when
$k \geq 1999$.

In other words, the sum of the first $k$ terms is less than 1.499 for $k=1,2,3,4$ as well as for $5 \leq k \leq 1998$, which is the same as
saying that this is true for $1 $\leq k \leq$
1998$.

Therefore, $k=1998$ is the largest
positive integer for which the sum of the first $k$ terms is less than 1.499.