Proving the arc length of a certain function is $\infty$ I'm having trouble with a homework problem. We're supposed to prove the following curve has arc length $\infty$ on the interval $[0,1]$:
\begin{gather*} 
f: \mathbb{R} \to \mathbb{R^2}: t \to 
\begin{cases}
(0,0) & \text{when t= 0}\\
(t, t\sin(\frac{\pi}{t})) & \text{otherwise}
\end{cases}
\end{gather*}
I was told by the lecturer that we are obliged to do this using the following formula:
\begin{gather*} 
L(f|_{[0,1]})= \sup\limits_{\mathcal{N} net}L(f|_{[0,1]}, \mathcal{N}) 
\end{gather*}
The net I came up with is $t_{j} = \frac{2}{3+2j}$. This seems right to me intuitively, but I can't seem to find a lower limit for it. I'd greatly appreciate some help with that.
 A: Let $t_n = \frac{2}{2n+1} > 0 $. This sequence is strictly decreasing and $t_1 = \frac{2}{3} < 1$. We have
$$\sin(\frac{\pi}{t_n}) = \sin(n\pi + \frac{\pi}{2})  = (-1)^n = \begin{cases} 1 & n \text{ even} \\ -1  & n \text{ odd}  \end{cases},$$
$$t_n  + t_{n+1} = \frac{2}{2n+1} + \frac{2}{2n+3} > \frac{1}{n} .$$
For each $k \in \mathbb N$ let
$$\mathcal N_k = \{s_0 = 0,s_1 = t_k, s_2 = t_{k-1}, \ldots, s_k = t_1, s_{k+1} = 1\}. $$
Then
$$L(f|_{[0,1]}, \mathcal{N}_k) =  \sum_{j=0}^k \lVert f((s_{j+1}) - f(s_j)) \rVert > \sum_{j=1}^{k-1} \lVert f((s_{j+1}) - f(s_j)) \rVert = \sum_{i=1}^{k-1} \lVert f(t_i) - f(t_{i+1}) \rVert .$$
For $i = 1,\ldots,k-1$ we have
$$f(t_i) - f(t_{i+1}) = (t_i,t_i\sin(\frac{\pi}{t_i})) - (t_{i+1},t_{i+1}\sin(\frac{\pi}{t_{i+1}})) = (t_i,t_i(-1)^i) - (t_{i+1},t_{i+1}(-1)^{{i+1}})) = (t_i -t_{i+1},(-1)^i(t_i + t_{i+1})) ,$$
thus
$$\lVert f(t_i) - f(t_{i+1}) \rVert = \sqrt{(t_i -t_{i+1})^2 + ((-1)^i(t_i + t_{i+1}))^2} = \sqrt{(t_i -t_{i+1})^2 + (t_i + t_{i+1})^2} >  \sqrt{(t_i + t_{i+1})^2}  = t_i + t_{i+1} > \frac{1}{i} .$$
Therefore
$$L(f|_{[0,1]}, \mathcal{N}_k) > \sum_{i=1}^{k-1}\frac{1}{i} .$$
But it is well-known that $\sum_{i=1}^{k-1}\frac{1}{i} \to \infty$ as $k \to \infty$ (harmonic series).
