Let $f:[0, 1] \to \mathbb{R}$ be an increasing function with $f(0) = 0$ and, for all $n \in \mathbb{Z}^{+}$, $f=1/n$ on $(\frac{1}{n+1},\frac{1}{n}].$ Find $$\int^{1}_{0} f dx$$ to one decimal place.
I've tried doing this by creating dissections but I cant see a way of doing it with all dissections being the same width, since the function goes up in steps, each time getting longer.
I then thought maybe they didn't have to be the same and tried $D_n = (0,\frac{1}{2n-1}, \frac{1}{2n-2}, ..., 1)$, but couldn't find the supremum and infimum for each of the dissections.
I'm aiming to use that if $U_D(f) - L_D(f) \to 0$, that means $$\int^{b}_{a} f = \lim L_D(f) = \lim U_D(f)$$
Break the integral up as follows:
$$\int_0^1 f \ dx = \int_{1/2}^1 f \ dx + \int_{1/3}^{1/2} f \ dx + \int_{1/4}^{1/3} f \ dx + \cdots = \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} f \ dx.$$
Now, on the interval $x \in (1/(n+1), 1/n]$, $f(x) = 1/n$, so
$$\sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} f \ dx = \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} \frac1n \ dx = \sum_{n=1}^\infty \frac1n \left(\frac1n - \frac1{n+1}\right) = \sum_{n=1}^\infty \frac1{n^2} - \sum_{n=1}^\infty \frac1{n(n+1)}.$$
The first sum on the right hand side is a well-known value,
$$\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6.$$
The second sum is a telescoping series,
$$\sum_{n=1}^\infty \frac1{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{n} -\frac1{(n+1)}\right) = 1.$$
Hence, the integral is equal to
$$\int_0^1 f\ dx = \frac{\pi^2}6 - 1.$$
You can approximate this as $\approx 0.645$