An example of a non-rectifiable curve How can I solve this problem?

Let the curve in $\mathbb{R}^2$ of the equation $\alpha(t)=(t,g(t)), t\in [0,1]$, where $$g(t)=\left\{ \begin{aligned} t \cos\left( \frac{\pi}{2t}\right), \quad t\not=0 \\ 0, \quad t=0\end{aligned}\right.$$
For $n \in \mathbb{N}$, let the partition $P$ of $[0,1]$ given by $$P=\left\{ 0, \frac{1}{2n},\frac{1}{2n-1},\ldots, \frac{1}{3},\frac{1}{2},1\right\},$$ prove the length $\ell_n$ of the polygonal inscribed satisfies $$\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$$and and deduce that the curve $\alpha$ is not rectifiable.

My approach:  I know for example that if I prove that arc length of $\alpha$ is infinite, so I can say that $\alpha$ is not rectifiable, so I need to calculate $I_{n}$.
I know that the arc-length of a curve $\alpha$ is $$s(t)=\int_{t_0}^t \|\dot{\alpha}(t)\| \, dt$$
so I need to prove that $s(t)\to \infty \implies \alpha$ is not rectifiable.
I know how calculate $\dot{\alpha}(t)$, but how can I choose $t_0$ and $t$?

Questions:
Now, How can I prove the first part? It's to say, how can I prove that $\ell_n>\sum_{k=1}^{2n}\frac{1}{k}$? and how can I relate that result to the fact that alpha is not rectifiable?
 A: For $n \in \mathbb N$, you have
$$\begin{aligned}\left\Vert \alpha(\frac{1}{n+1}) - \alpha(\frac{1}{n}) \right\Vert &=\sqrt{\left(\frac{1}{n+1}-\frac{1}{n}\right)^2+\left(\frac{\cos\left(\frac{\pi(n+1)}{2}\right)}{n+1}-\frac{\cos\left(\frac{\pi n}{2}\right)}{n}\right)^2}\\
&\ge \frac{1}{n+1}
\end{aligned}$$
as one of $\cos\left(\frac{\pi(n+1)}{2}\right) , \cos\left(\frac{\pi n}{2}\right)$ vanishes while the absolute value of the other one is equal to one. This enables to get the required inequality
$$l_{n}>\sum_{k=1}^{2n}\frac{1}{k}.$$
As the arc length of  a curve is greater than a polygonal path approximating it and that the harmonic series $\sum \frac{1}{n}$diverges, you get that $\alpha$ is not rectifiable.
A: Here is the idea. In the euclidean plane, the shortest path between two point is the straight line. Let $P = \{ 0,\frac{1}{2n}, \frac{1}{2n-1},\ldots, \frac{1}{2},1\} = \{ x_0,x_1,\ldots x_{2n}\}$. Consider the points $M_k$ on the graph of $g$ defined by $M_k = (x_k,g(x_k))$. Because the straight line is the shortest path in the plane, one has $\mathrm{length}(g|_{[x_k,x_{k+1}]}) \geqslant \mathrm{length}([M_k,M_{k+1}])$. As $P$ is a partition of $[0,1]$, $\mathrm{length}(g|_{[0,1]})=\sum_{k=0}^{2n-1}\mathrm{length}(g|_{[M_k,M_{k+1}]})$.
So now you have the inequality:
\begin{align}
\forall n \geqslant 1,~ \mathrm{length}(g) \geqslant A_n
\end{align}
where $A_n = \sum_{k=0}^{2n-1} \mathrm{length}\left([M_k,M_{k+1}]\right)$. Try to bound from below these last terms and you are done.
