We look at approximation by integrals today.When a summation is expressed in terms of a function of k where this f(k) is a monitonically increasing function, we can approximate it by integrals because it finds the area under the curves in slices. By using the summation of the rectangles under the curve we can get a pretty good idea on the bounds of the function. Moreover, this applies to both the monotonically increasing and decreasing curves.
In the case of the monotonically increasing function on a graph from left to right, the area of the slice on the left of a chosen slice will be lesser or equal and the slice on the right of the chosen slice will be higher or equal. That is we have a generic three partitions of the ranges and we can show this for any of the repeating three slices.
The integral approximation gives a tight estimate for the nth harmonic number. For a lower bound, we obtain.
Sum of k = 1 to n of (1/k)  greater than or equal to Integral of 1 to n slices of (1/x) dx = ln (n  + 1) because each slice can be of unit width.
For the upper bound we derive the inequality
Sum of k = 2 to n of (1/k) is less than or equal to Integral of slices (1/x)(dx) = ln (n) again based on  unit-width slices
Together this yields the bound of the harmonic series Sum of k  = 1 to n of (1/k) <= ln (n)  + 1.
 We note that the total area under the curve is the the value of the summation. The integral is represented by the shaded area under the curve. By comparing the areas for the lower and upper bound, we see that the rectangles are slightly under the curve in the case of the lower bound and slightly over the curve  in the case of the upper bound. We compute the areas as
Sum m-1 to n of f(x)dx  <= Sum m to n of f(x)dx and by shifting the rectanges one to the right we establish sum m to n of f(x)dx <= Sum m to n+1 of f(x)dx.