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

Question

Elijah Abram · Accepted Answer

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:

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.

Axel Alvarez · Answer

undefined To use the integral test to determine if the series $ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \ldots $ is convergent or divergent, follow these steps:

1. Consider the function $ f(x) = \frac{1}{x} $. This function is positive, continuous, and decreasing for $ x \geq 3 $, which includes all terms in the series.

2. Integrate $ f(x) $ from 3 to infinity:
   $$ \int_{3}^{\infty} \frac{1}{x} \, dx = \ln(x) \bigg|_{3}^{\infty} = \lim_{b \to \infty} (\ln(b) - \ln(3)) $$

3. Evaluate the limit:
   $$ \lim_{b \to \infty} (\ln(b) - \ln(3)) = \infty $$

4. Since the integral diverges, by the integral test, the series $ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \ldots $ also diverges.