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

1 Answer
Nov 23, 2016

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.