Problem 530

We note that $t_{1}=\frac{1}{1}-\frac{1}{3}=\frac{2}{3} \approx 0.67$, $t_{1}+t_{2}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{2}{3}+\frac{1}{4}=\frac{11}{12} \approx 0.92$, $t_{1}+t_{2}+t_{3}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}=\frac{1}{1}+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}=1.05$, $t_{1}+t_{2}+t_{3}+t_{4}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}=\frac{1}{1}+\frac{1}{2}-\frac{1}{5}-\frac{1}{6} \approx 1.13$. 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 $t_{1}+t_{2}+t_{3}+\ldots+t_{k-1}+t_{k}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{k-1}-\frac{1}{k+1}\right)+\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k-1}+\frac{1}{k}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\cdots-\frac{1}{k+1}-\frac{1}{k+2}=\frac{1}{1}+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}=1.500-\frac{1}{k+1}-\frac{1}{k+2}$. This means that the sum of the first $k$ terms is less than 1.499 exactly when $\frac{1}{k+1}+\frac{1}{k+2}$ is greater than 0.001. As $k$ increases from 4, each of $\frac{1}{k+1}$ and $\frac{1}{k+2}$ decreases, which means that their sum decreases as well. When $k=1998, \frac{1}{k+1}+\frac{1}{k+2}=\frac{1}{1999}+\frac{1}{2000}>\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}=0.001$. When $k=1999, \frac{1}{k+1}+\frac{1}{k+2}=\frac{1}{2000}+\frac{1}{2001}<\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}=0.001$. This means that $\frac{1}{k+1}+\frac{1}{k+2}$ 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.