Question

How do you use the integral test to determine if `1/3+1/5+1/7+1/9+1/11+...` is convergent or divergent?

Answer 1

First, we have to write a rule for this summation. Note the denominator is increasing by `2` each time. This summation works:

`sum_(n=0)^oo1/(2n+3)`

There are other summations that would work here as well, but this will suffice.

In order for `sum_(n=N)^oof(n)` to be testable for the integral test, it must fit two conditions:

• `f(n)` must be positive
• `f(n)` must be decreasing

Both of these are true: all the terms are greater than `0` and as `n` increases, the denominator increases, so the terms as wholes decrease. Thus the integral test will apply here.

The integral test states that for `sum_(n=N)^oof(n)` where `f(n)` fits the criteria, if `int_N^oof(x)dx` converges, that is, is equal to a value, then `sum_(n=N)^oof(n)` converges as well.

So, we will evaluate the following integral:

`int_0^oo1/(2x+3)dx=lim_(brarroo)int_0^b1/(2x+3)dx`

`color(white)(int_0^oo1/(2x+3)dx)=1/2lim_(brarroo)int_0^b2/(2x+3)dx`

`color(white)(int_0^oo1/(2x+3)dx)=1/2lim_(brarroo)[ln(|2x+3|)]_0^b`

`color(white)(int_0^oo1/(2x+3)dx)=1/2lim_(brarroo)(ln(2b+3)-ln(3))`

As `brarroo`, `ln(2b+3)rarroo` as well.

`color(white)(int_0^oo1/(2x+3)dx)=oo`

The integral diverges, so `sum_(n=0)^oo1/(2n+3)` diverges also.