I am trying to solve the following problem:
Consider the curve $$ \begin{align*} f(t) &= (t, \cos(\frac{\pi}{t}) \textrm{ whenever } t\neq 0 \\ f(0) &= (0, 0) \end{align*} $$ Let $P$ be the partition $$P = \{0, \frac{1}{n}, \frac{1}{n-1}, ...,\frac{1}{2}, 1\}$$ Show that the inscribed polygon $\pi(P)$ has length: $$ |\pi(P)| \geq 1 + 2(\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) $$
I've tried to set up the length a few different ways, but I haven't gotten any expressions that have the form from above.
What I have so far is: let $P_k = \frac{1}{k}$ for $1 \leq k \leq n$, and ignore the point at 0 for now. Then the length of the $k$th segment is given by $|| f(P_{k+1}) - f(P_k) ||$. Doing some algebra, and taking advantage of the fact that $\cos((\pi + 1)k) = -\cos(\pi k)$:
$$ \begin{align*} f(P_k) &= (\frac{1}{k}, \frac{\cos(\pi k)}{k})\\ f(P_{k+1}) - f(Pk) &= (\frac{1}{k+1} - \frac{1}{k}, \frac{\cos(\pi(k + 1)}{k+1} - \frac{\cos(\pi k)}{k})\\ &= (\frac{-1}{k^2 + k}, \frac{\cos(\pi(k+1))}{k + 1} + \frac{\cos(\pi(k+1))}{k} \\ &= (\frac{-1}{k^2 + k}, \frac{-(2k +1)\cos(\pi k)}{k^2 + k}) \end{align*} $$
So taking the norm I end up with $$\sqrt{\frac{1}{(k^2 + k)^2} + \Big(\frac{(2k + 1)\cos(\pi k)}{k^2 + k}\Big)^2}$$
and since $\cos(\pi k)$ is always either $1$ or $-1$, it becomes: $$\sqrt{\frac{1}{(k^2 + k)^2} + \Big(\frac{2k + 1}{k^2 + k}\Big)^2} = \sqrt{\frac{(2k+1)^2 + 1}{(k^2+k)^2}}$$
and the first few terms of the sum are $\sqrt{\frac{5}{2}}$, $\frac{1}{3} \sqrt{\frac{13}{2}},...$, which does not match what I'm expecting, and seems to be a bit numerically smaller.
Where have I gone wrong here?