Show that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on $[2^{-2n} , 2^{-2n+2}]$ for any positive integer $n$.
So that readers may be able to give precise feedback as to where my understanding is erroneous, I will lay out my calculations and reasoning for each step.
My Work
I know that the formula for the Riemann sum is $lim_{n \to \infty} \sum_{j = 1}^n f(c_j)\Delta x_j$.
$c_j$ is some element in the interval of integration $[x_{j-1}, x_j]$. Since we're taking the Riemann sum ($n \to \infty$ and the norm/magnitude of each partition goes to $0$), the value of $x$ in the interval of integration for which we take the function value at does not matter.
$\Delta x_j$ represents the magnitude/norm of each partition. It is calculated using $\dfrac{b - a}{n} = \Delta x_j$ where $b$ and $a$ are the bounds of integration.
For this problem, the interval/domain of integration is $[2^{-2n}, 2^{-2n+2}]$ for all positive integers $n$.
Therefore, we can calculate the norm/magnitude of this partition (for any positive integer $n$) as $2^{-2n+2} - 2^{-2n} = 3 \cdot 2^{-2n}.$ This is our $\Delta x_j$.
For $c_j$, we can select any point in the interval $[2^{-2n}, 2^{-2n+2}]$. I will select $c_j = c_n = 2^{-2n}$.
Therefore, the Riemann sum is $\sum_{n = 1}^\infty f(2^{-2n}) \cdot 3 \cdot 2^{-2n} = 3 \cdot \sum_{n = 1}^\infty \dfrac{1}{\sqrt{\dfrac{1}{2^{2n}}}} \cdot 2^{-2n}$
I have no idea how to continue from here.
I would greatly appreciate it if people could please take the time to review my work and provide feedback.
