Let $f(t) = 1/t^2$ and suppose want to approximate $\int_1^{\infty}f(t) dt$ as a Riemann sum.
Suppose we use rectangles of equal width, say $x$. Then the edges of the rectangles occur at the points $t= 1, 1+x, 1+2x, 1+3x, \ldots$. We can express this concisely as a sequence $t_n = 1+nx$, where $n=0,1,2,\ldots$.
Let us consider two approximations.
Approximation from below
Here, we choose the height of each rectangle as tall as possible so that the rectangle remains entirely below the graph of the function. Since $f$ is decreasing, this means that we choose the height of the rectangle so that its upper right corner touches the function. So, the height of the first rectangle will be $f(t_1)$, the height of the second rectangle will be $f(t_2)$, etc.
Then the sum of the areas of these rectangles is
$$\sum_{n=1}^{\infty}f(t_n)x$$
since $f(t_n)$ is the height of the $n$th rectangle, and every rectangle has width $x$. Since $f(t_n) = f(1+nx) = 1/(1+nx)^2$, this becomes
$$\sum_{n=1}^{\infty}\frac{x}{(1+nx)^2}$$
and we would expect that as we make the width $x$ of the rectangles small, this should converge to the integral of $f$:
$$\lim_{x \to 0}\sum_{n=1}^{\infty}\frac{x}{(1+nx)^2} = \int_1^{\infty}\frac{1}{t^2}dt$$
Approximation from above
In this case, we choose the height of each rectangle as short as possible so that the rectangle remains above the graph of the function. Since $f$ is decreasing, this means that we choose the height of the rectangle so that its upper left corner touches the function. So, the height of the first rectangle will be $f(t_0)$, the height of the second rectangle will be $f(t_1)$, etc.
Then the sum of the areas of these rectangles is
$$\sum_{n=0}^{\infty}f(t_n)x$$
Since $f(t_n) = f(1+nx) = 1/(1+nx)^2$, this becomes
$$\sum_{n=0}^{\infty}\frac{x}{(1+nx)^2}$$
Notice that the only difference versus approximation from below is that the sum starts at $n=0$ instead of $n=1$.
As before, if we make the width $x$ small, we would expect that in the limit, we should get the integral:
$$\lim_{x \to 0}\sum_{n=0}^{\infty}\frac{x}{(1+nx)^2} = \int_1^{\infty}\frac{1}{t^2} dt$$
Now let's check that the above makes sense in terms of convergence. Let us define $L(x)$ and $U(x)$ to be the lower and upper approximations (i.e., the approximations from below and above, respectively). Therefore,
$$L(x) = \sum_{n=1}^{\infty}\frac{x}{(1+nx)^2}$$
and
$$U(x) = \sum_{n=0}^{\infty}\frac{x}{(1+nx)^2}$$
It's easy to see that for each fixed $x \geq 0$, both sums converge absolutely, so $L(x)$ and $U(x)$ are well defined. Moreover, their difference is simply the $n=0$ term:
$$U(x) - L(x) = \frac{x}{(1+0x)^2} = x$$
Therefore, as $x \to 0$, we have
$$\lim_{x \to 0}\ (U(x) - L(x)) = \lim_{x \to 0}\ x = 0$$
which shows that the difference between the upper and lower approximations converges to zero as we shrink the rectangle widths to zero. Since the integral $\int_1^{\infty}(1/t^2)dt$ is sandwiched between these two approximations, both $U(x)$ and $L(x)$ converge to $\int_1^{\infty}(1/t^2)dt = 1$ as $x \to 0$.
