When using the Integral test, why is the value of the integral different from the sum of the series?

Question

According to my textbook, the value of the improper integral is not always equal to the sum of the series. But why is that?

Answer 1

A series is like adding up a bunch of actual rectangles. You can see from this image that there is a difference in the areas of the rectangles and the area under the smooth curve. The darker orange part is how much the sum is greater than the integral. In this case the sum diverges but the points is the same.

Answer 2

From an "algebraic" perspective, note that we can construct the improper integral $\int_1^\infty f(i)di$ as we partition the limits of integration $[1,\infty)$ into infinite intervals with constant lenght $a$, define rectangles whose lenght is $a$ and individual height is the function of some value in the respective interval, and take the limit when $a \to 0$, i.e, a Riemann sum:

$$\int_1^\infty f(i)di = \lim_{a \to 0} \sum_{j=0}^{\infty} a f(aj+1).$$

Now, realize that if $a=1$, we have $\sum_{j=0}^{\infty} f(j+1)$, which is exactly our original infinite sum. However, for different values of $a$ we don't have necessarily equal values of the sum. For example, if the function $f$ has the property $(\forall x,y \in Dom(f))(x>y \implies f(x)>f(y))$, then $a>b \implies \sum_{j=0}^{\infty} a f(aj+1)> \sum_{j=0}^{\infty} b f(bj+1)$. Also remember that the integral is the limit of the sum when $a \to 0$, hence in the above case the integral is smaller than the series.