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

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.