The exercise is to show that $f(t) = t \cos \frac{\pi}{2t}$ is not rectifiable.
To show that, Tom Apostol is guiding us to consider the partition $P = \{0, \frac{1}{2n}, \frac{1}{2n-1}, ..., \frac{1}{3}, \frac{1}{2}, 1\}$, and to show the corresponding inscribed polygon $\pi(P)$ has length greater than $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}$.
Where did he pull $1$ from?
The only explanation I could get is that he connected the end dots of the polygon, but that's not how he would normally consider an inscribed polygon (e.g. in section $14.10$, and figures $14.12$ or $14.14$).
Thanks!
Yes, the $1$ seems to be wrong, but it's irrelevant, as we are bounding the length below by a divergent series, regardless.