### Video Transcript

What can we conclude by applying
the πth term divergence test in the series the sum of three π over the square root
of six π squared plus four π plus five for values of π from one to infinity?

Remember, the πth term test for
divergence says that if the limit as π approaches infinity of ππ does not exist
or the limit is not equal to zero, the series the sum of ππ from π equals one to
infinity is divergent. And we recall that if the limit is
equal to zero, we canβt tell whether the series converges or diverges and we say
that the test fails.

In this question, weβre going to
let ππ be equal to three π over the square root of six π squared plus four π
plus five. And so we need to evaluate the
limit of this expression as π approaches infinity. We canβt use direct
substitution. If we were to do so, we would get
infinity over infinity, which we know is indeterminate. So instead, we need to find a way
to manipulate our expression and see if that will help us to evaluate the limit.

To make our life just a little bit
easier, letβs apply the constant multiple rule. And this is that the limit as π
approaches infinity of some constant times the function in π is equal to that
constant times the limit of the function in π. So essentially, we can take the
constant three outside of our limit.

And then this next step is going to
appear a little bit strange. Weβre going to divide both the
numerator and denominator of our expression by π. The numerator then becomes one. And the denominator becomes one
over π times the square root of six π squared plus four π plus five. Weβre now going to take this one
over π inside our square root. To do that, weβll need to square
it. So the denominator becomes the
square root of six π squared over π squared plus four π over π squared plus five
over π squared. And then, actually, this simplifies
quite nicely.

We now have three times the limit
as π approaches infinity of one over the square root of six plus four over π plus
five over π squared. And weβre now ready to apply direct
substitution. As π gets larger, four over π and
five over π squared get smaller. They approach zero. One and six are both independent of
π. So our limit becomes one over the
square root of six.

We want to rationalise the
denominator here. So we multiply both the numerator
and denominator of our fraction by the square root of six. And we see that the limit as π
approaches infinity of ππ is three times the square root of six over six, which is
root six over two. We notice that this is not equal to
zero. And this leads us to conclude that
the series the sum of three π over the square root of six π squared plus four π
plus five between π equals one and infinity is divergent.