We want to determine
if the given series converges or diverges.
To help us decide on an appropriate test,
let's take a look at our infinite series.
Notice how we have the sum of one divided by
the square root of the quantity N cubed plus three N.
Notice, if we ignore this plus three N,
this series would resemble one divided by
the square root of N cubed,
which if we wrote using a rational exponent,
would be one divided by N to the three halves,
which would be a known converging series
by the P-Series test.
Again, if we use the fact that the Nth root
of A to the power of M is equal to
A to the power of M divided by N,
the given series resembles the sum of
one divided by N to the three halves,
which we know converges by the P-Series test,
with P equal to three halves, which is greater than one.
If we have a series that resembles
a known converging or diverging series,
the two most appropriate test will be the
limit comparison test, or the direct comparison test.
Remember, in some cases, more than one test will work.
I think, for this example, we will go through both tests,
just to show that more than one test can be used.
Let's start by reviewing
and applying the limit comparison test.
The limit comparison test tells us that if the limit
as N approaches infinity of A sub N
divided by B sub N equals L, which is greater than zero,
then if the sum of B sub N converges,
then so does the sum of A sub N,
and if the limit as N approaches infinity of A sub N
divided by B sub N equals L, which is greater than zero,
where the sum of B sub N diverges,
then so does the sum of A sub N.
Because we're comparing to a known converging series,
we want to show this limit here is greater than zero,
where the sum of B sub N is a known converging series,
and the sum of A sub N is the given series.
We want to find the limit as N approaches infinity
of A sub N, which is going to be from the given series.
We will have one divided by,
I'm going to write this as the quantity
N to the third plus three N to the one half power,
instead of the square root,
and we will divide this by B sub N,
which is from the known converging series
of one divided by N to the three halves.
We have the limit of this complex fraction,
where this division bar represents
the division of two fractions,
which we are going to write is a multiplication problem.
Instead of dividing by the bottom fraction,
we'll multiply by the reciprocal.
This is going to be equal to the limit
as N approaches infinity of the top fraction,
and then times the reciprocal of the bottom fraction,
which would be N to the three halves divided by one.
Nothing simplifies here,
so we will have the limit as N approaches infinity
of N to the three halves,
divided by the quantity N cubed
plus three N to the one half power.
Now again, notice in the denominator,
if we ignore this plus three N,
because as N approaches infinity,
only N to the third matters,
the degree of the denominator would be three halves,
which is the same as the degree of the numerator,
and therefore, this limit is equal to the ratio
of the leading coefficients,
which would just be one divided by one,
which is equal to one.
Notice, this limit is greater than zero, and therefore,
since we compared it to a known converging series,
the given series also converges.
Buy limit comparison test,
with the sum of B sub N equals
the sum of one divided by N to the three halves,
which converges by the P-Series test,
and the sum of A sub N equals the given series,
since the limit as N approaches infinity of A sub N
divided by B sub N equals one, which is greater than zero,
the given series converges.
Now, let's also see if we can determine that
the given series converges,
using the direct comparison test.
Let's first review the direct comparison test.
Given the sum of A sub N, and B sub N, such that
A sub N is greater than zero,
and less than or equal to B sub N,
then, if the sum of B sub N converges,
then so does the sum of A sub N,
and if the sum of A sub N diverges,
then the sum of B sub N also diverges.
Because we're trying to show convergence,
we want to show this statement here is true,
where B sub N is a known converging series,
so we want to show that the terms of the given series,
from A sub N, will always be less than or equal to B sub N,
the terms from the known converting series.
If these terms are less than these
terms from the converging series,
we can conclude that the sum of A sub N also converges.
From the limit comparison test that we just did,
we know we are going to compare to the sum
of one divided by N to the three halves,
which we know converges.
We are going to let this equal the sum of B sub N,
and therefore, the given series is going to be
the sum of A sub N.
By the direct comparison test,
since the sum of one divided by N to the three halves
converges by the P-Series test,
and we want to show that zero is greater than A sub N,
which is less than or equal to B sub N, or in this case,
zero is less than A sub N is from the given series,
which I'm going to write to the one half power,
instead of the square root,
we want to show this is less than or equal to B sub N,
which is one divided by N two three halves.
If this is true, we can conclude that the given series is
also convergent using the direct comparison test.
Looking at these values,
notice how this denominator here
is going to have a larger denominator,
because of the plus three N,
and since the numerators are the same,
the fraction with the larger denominator
is going to be smaller than
the fraction with the larger denominator.
To get a better feel for this,
let's take a look at these two fraction values
on the graphing calculator.
To save some time, I've already typed in these fractions.
In Y one, I entered one divided by the quantity
X to the third plus three X raised to the one half power.
Notice how I used X instead of N on the graphing calculator.
Then, for Y sub two,
I entered one divided by X to the power of three halves.
Now, if we go to the table, by pressing second graph,
notice how Y sub one is always less than or equal
to Y sub two, therefore this compound inequality is true,
and therefore, by the direct comparison test,
we can conclude that the given series is convergent.
Notice how as X, or in this case N, increases,
These values are very close together,
but notice how Y sub one is always less than Y sub two.
To complete our statement here,
since this condition is met,
the given series converges.
Okay, I hope you found this helpful.