We're going to determine if the infinite series
converge or diverge.
Looking at the two series,
notice how they're almost the same except
this series alternates
and this series doesn't.
So to determine if this series converges or diverges,
I think we'll apply the limit comparison test.
So to begin,
we'll find an infinite series that resembles this series
that we know converges or diverges.
Notice how the given series resembles the series
where we'd have the summation
from n equals one to infinity of n to the fifth
divided by the square root
of n to the 11th.
But we know the square root of n to the 11th
is equal to n to the 11/2.
So let's write this as the summation
from n equals one to infinity of
n to the fifth divided by n to the 11/2.
And now because the bases are the same and we're dividing,
we'll simplify this by subtracting the exponents.
n to the fifth divided by n to the 11/2
would be equal to n to the power of five minus 11/2,
which would be 10/2 minus 11/2 or negative 1/2,
which you can write as one divided by n to the 1/2.
So we can write this as the summation,
from n equals one to infinity,
of one divided by n to the 1/2,
which we should recognize would diverge
by the p-Series test.
So for a quick review of the limit comparison test,
since our series, the summation of a sub n,
resembles the summation of b sub n
which we know diverges,
if the limit of a sub n divided by b sub n equals L,
where L is positive and finite,
then since we know the summation of b sub n diverges,
so does the summation of a sub n.
So we'll be trying to show this series diverges
but a limit comparison test.
So now we'll take the limit,
as n approaches infinity,
of a sub n which is n to the fifth
divided by the square root
of n to the 11th plus three
divided by b sub n.
Let's write b sub n in this form here,
so we have n to the fifth
divided by the square root of n to the 11th.
And now we'll write this quotient as a product.
This is equal to the limit,
as n approaches infinity,
of the fraction in the numerator.
And then we'll multiply it by the reciprocal
of the fraction, the denominator,
which would be times the square root of n to the 11th
divided by n to the fifth.
Notice how in this form
n to the fifth over n to the fifth simplifies to one.
So now we have the limit,
as n approaches infinity, of
the square root of n to the 11th which I'll now write as
n to the 11/2
divided by the square root
of n to the 11th plus three.
Now, in this form we can take a shortcut
to determine this limit.
But look at the denominator and notice how the plus three
is irrelevant as n approaches infinity.
So we can think of the denominator as
the square root of n to the 11th
[mumbles] would be equal to n to the power of 11/2.
In this case, we can think of the numerator and denominator
of having the same degree,
and therefore, the limit would be equal to the
ratio of the leading coefficients,
which would be one over one or just one.
So because this limit is positive and finite,
we know the given series also diverges.
But if we do want to show work on this,
we would divide everything by the highest power of n
in the denominator
which would be n to the 11th under the square root.
So if we do want to show work, we can write this as the limit
as approaches infinity of,
in the denominator we'd have the square root of
n to the 11th divided by n to the 11th
plus three
divided by n to the 11th.
But n to the 11th under the square root
is equal to n to the 11/2 outside the square root.
So we'd divide the numerator by n to the 11/2.
So we'd have n to the 11/2
divided by n to the 11/2.
Now, if we simplify this,
we have the limit as n approaches infinity of,
this would be one
divided by
the square root of one plus three
divided by n to the 11th.
In this form, notice how
we can more easily see the limit is equal to one.
And because this is positive and finite
and we compared the given series to a diverging series,
the given series also diverges.
So let's go ahead and summarize the results
and then we'll take a look at the second infinite series.
We compared the given series to
the summation from n equals one to infinity of
n to the fifth divided by the square root of n to the 11th,
which is equal to
the summation of n equals one to infinity of
one divided by n to the 1/2,
which diverges by the p-Series test
with p equals 1/2
which is less than or equal to one.
And then by the limit comparison test,
since the limit
as n approaches infinity of a sub n divided by b sub n
equals one,
which is positive and finite,
the given series also diverges.
Now, let's take a look at our second series,
the alternating series.
We'll be applying the alternating series test
to determine convergence or divergence.
So first, notice how a sub n,
the non-alternating part,
is equal to n to the fifth
divided by the square root of n to the 11th plus three.
Notice how, here, a sub n is always greater than zero
for all values of n.
And now to apply the alternating series test,
we need to show the limit as n approaches infinity
of a sub n equals zero
and a sub n plus one
is less than or equal to a sub n.
So we'll first take the limit,
as n approaches infinity,
of n to the fifth
divided by the square root of n to the 11th plus three.
And, again, there's a shortcut for determining this limit.
We can think of the denominator as just
the square root of n to the 11th,
which would be equal to n to the power of 11/2
which is equal to 5.5.
So we can think of the denominator
as having a higher degree than the numerator,
and therefore, this limit will equal zero.
But, again, if we do want to show work,
we divide everything by the square root of n to the 11th,
which would be the limit,
as n approaches infinity, of,
in the denominator we'd have the square root of
n to the 11th
divided by n to the 11th
plus three divided by
n to the 11th.
And then for the numerator,
we would divide n to the fifth
by n to the 11/2.
So notice how here we have the limit,
as n approaches infinity of,
this would be one divided by n to the 1/2.
The denominator would be the square root of one plus three
divided by n to the 11/2.
And notice how as n approaches infinity,
the numerator approaches zero,
this fraction approaches zero,
this stays at one.
So this limit does equal zero.
Now, for the last step, we want to show that a sub n plus one
is less than or equal to a sub n.
To show this, look at several of the terms
using our formula here for a sub n.
I've already done this to save some time.
And here they are.
Notice here, the terms actually increase
from the first term to the second term.
But when n is greater than or equal to two,
the terms are getting smaller and smaller,
and, therefore, a sub n plus one is less than a sub n.
[mumbles] to make a note here,
this is only true when n is greater than or equal to two.
So let's go ahead and summarize the results.
By the alternating series test,
since,
let's go ahead and say that a sub n is greater than zero,
the limit as n approaches infinity of a sub n
equals zero, and
a sub n plus one is less than or equal to a sub n,
for n greater than or equal to two,
our alternating series
converges.
So notice how the alternating series converges,
but the non-alternating series diverges,
and therefore, we can say
the alternating series is conditionally convergent,
which we'll learn more about in our next lesson.
I hope you found this helpful.