We want to use the alternating series test
to determine if the given infinite series
converges or diverges
and what can we conclude.
So for a quick review, if we have an alternating series
in this form here, where a sub n is greater than zero,
the alternating series converges
if the limit as n approaches infinity
of a sub n equals zero
which is the nth term divergent test
and a sub n plus one is less than
or equal to a sub n for all n,
meaning the terms are non-increasing.
When we have an alternating series,
we can also test for what's called conditional
or absolute convergence which we'll discuss more
in a future lesson.
So going back to our example,
notice how a sub n, the non-alternating part,
would be equal to n to the fifth
divided by the square root of n to the ninth plus two.
Well, first notice that a sub n is always positive
for all values of n.
So now we'll determine the limit
as n approaches infinity of a sub n.
Now, there is a shortcut for determining this limit,
by looking at the degree of the numerator and denominator.
Looking at the denominator first,
notice how the plus two is irrelevant
as n approaches positive infinity.
And we can think of the denominator
as just the square root of n to the ninth.
And because the index is two, we can write this
as n raised to the power of nine halves.
So we can think of the denominator
as having degree nine halves
and the numerator has degree five.
So because the numerator has a higher degree,
as n approaches infinity, this quotient
also approaches positive infinity.
And since this limit is not equal to zero,
the alternating series test doesn't apply
and the alternating series diverges
by the nth term divergent test.
But if we do want to show some work here,
we would divide by the highest power of n
in the denominator
which would be n to the ninth under the square root.
So to show some work, we can write this as the limit
as n approaches infinity of.
In the denominator we'd have the square root
of n to the ninth divided by n to the ninth
plus two divided by n to the ninth.
And now for the numerator,
because the n to the ninth is underneath the square root,
we would divide the numerator by n to the nine halves.
So we'd have n to the fifth
divided by n to the nine halves.
And now, if we simplify, we would have the limit
as n approaches infinity of.
In the numerator, we'd have n to power
of five minus nine halves
or 10 halves minus nine halves which would be one half.
And our denominator would be the square root
of one plus two divided by n to the ninth.
Now in this form, notice how the denominator
would approach the square root of one or one
since this fraction approaches zero
as n approaches positive infinity.
And the numerator increases without bound
as n approaches infinity.
So therefore, the alternating series test
does not apply.
And because this limit doesn't equal zero,
we know the series diverges
by the nth term divergent test.
We could also say that the absolute value
of the terms doesn't approach zero as it increases.
And therefore, the series diverges.
But before we go, I do want to look at this last condition
that says a sub n plus one must be less than
or equal to a sub n.
Well, notice a sub one would be equal
to one to the fifth divided by the square root
of one to the ninth plus two
which would be approximately 0.577.
a sub two would be equal to two to the fifth
divided by the square root of two to the ninth plus two.
This is approximately 1.992.
Notice how the terms are actually increasing.
And let's look at one more term.
a sub three would be three to the fifth
divided by the square root of three to the ninth plus two.
And this comes out to approximately 3.000.
So to see how this last condition is also not true,
a sub n plus one is not less than
or equal to a sub n.
But we know the series diverges
because it fails the nth term divergent test.
I hope you found this helpful.