A Nonrectifiable Curve Do Carmo's book Page 11.
Let $\alpha : [0,1] \to \mathbb{R}^2$ be given as $$\alpha(t)=
\begin{cases} 
(t,t \sin (\pi/t)) & \text{ for } t \ne 0\\
(0,0) & \text{ for } t=0 \,.
\end{cases}$$ Show that the arc length of the portion of the curve corresponding to $\dfrac{1}{n+1} \leq t \leq \dfrac{1}{n}$ is at least $2/(n+1/2)$. Use this to show that the length of the curve in the interval $1/N \leq t \leq 1$ is greater than $2 \sum_{n=1}^N 1/ (n+1)$.
For the first part, I considered the distance between $\bigl(\alpha(1/n), \alpha(2/(2n+1))\bigr)$ and $\bigl(\alpha(2/(2n+1)), \alpha(1/(n+1))\bigr)$ and proved that it should be at least $2/(n+1/2)$.
In the second part, we can see that
\begin{align}
[1/N, 1] & = [1/N, 1/(N-1)]\cup [1/(N-1), 1/(N-2)]\cup \dots \cup [1/2,1] \,.
\end{align}
Therefore,
\begin{align}
|\alpha(1)-\alpha(1/2)|+ \dots + |\alpha(1/(N-1))-\alpha(1/N)| & \geq \sum_{n=1}^{N-1}\dfrac{2}{n+1/2} \,,
\end{align} and that won't give the desired solution. I don't know where I have made a mistake.
 A: Everything you have is correct. All that's remaining is to show that your lower bound is at least as large as the desired lower bound. Equivalently, we need to show that
$$\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{1}{n+1} > 0 \,.$$
We have the following.
\begin{align*}
\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{1}{n+1}
&= \sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^{N-1} \frac{1}{n+1} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \biggl(\frac{1}{n+0.5} - \frac{1}{n+1}\biggr) - \frac{1}{N+1} \\
&= \sum_1^{N-1} \biggl(\frac{n+1}{(n+0.5)(n+1)} - \frac{n+0.5}{(n+0.5)(n+1)}\biggr) - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{n+1-n-0.5}{(n+0.5)(n+1)} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{0.5}{(n+0.5)(n+1)} - \frac{1}{N+1} \\
&= \sum_1^{N-1} \frac{1}{(2n+1)(n+1)} - \frac{1}{N+1}
\end{align*}
The final term is approximately equal to $0.07$ for $N=4$ and is an increasing function in $N$, so the claim holds for all $N \geq 4$.
This doesn't necessarily show the statement is false for $N \leq 3$ but doesn't prove it true either.
