Welcome to another example on determining
whether an infinite series is converging or diverging.
We do have general guidelines outlined here
for determining if an infinite series converges or diverges.
However, I think what we'll find as we practice,
we'll recognize which test to apply
based upon the form of the infinite series.
Let's take a look at our example.
We want to determine if the given series
converges or diverges and state the test used.
Typically we want to start with the nth term divergent test,
which is a test to see if the nth term approaches zero.
This test tells us that if the limit
as n approaches infinity of a sub n doesn't equal zero,
we can quickly determine that the series is divergent.
Notice how if the limit does equal to zero,
this test doesn't tell us anything,
this series can still be divergent or convergent
based upon the outcome of a different test.
So looking at the given series,
notice that a sub n would be this quotient,
so we're going to find the limit
as n approaches infinity
of two divided by n times the cubed root of n.
This is a pretty straight-forward limit,
notice the numerator is always two
and the denominator increases without bound,
and therefore this fraction approaches zero,
and therefore this limit is equal to zero.
So again, when this limit is equal to zero,
the nth term divergent test doesn't tell us anything,
so let's go ahead and just state that
the nth term divergent test
is inconclusive for this series.
Which means we'll have to apply a different test
to determine if this is convergent or divergent.
To recognize which test to apply,
let's rewrite the denominator using rational exponents.
If we consider a sub n just for a moment,
it would be two divided by n times,
the cubed root of n is the same as
the cubed root of n to the first,
using our rational exponent rule here,
this should be n to the 1/3 power.
And n is the same as n to the first,
so we can write a sub n as two
divided by n to the 4/3 power.
Which means we can rewrite the given series
as the sum from n equals one to infinity
of two divided by n to the 4/3.
So in this form we should recognize
that the series does resemble the p-series.
So now we'll apply the p-series test.
So to review, the p-series test tells us
that the infinite series in this form
converges if p is greater than one
and diverges if p is greater than zero
and less than or equal to one.
And again, we just rewrote the given series as
the summation of n equals one to infinity
of two divided by n to the 4/3 power.
But it would also be helpful if we factor out this two
so it's an exact match to the p-series.
So we can write this as two times
summation from n equals one to infinity of one
divided by n to the 4/3.
Notice if this series here converges,
then two times a converging series would also converge.
And in this form, it's much easier to recognize
that p is equal to 4/3,
and 4/3 is greater than one,
and therefore this series will converge.
So by the p-series test, the given infinite series,
which is equal to two times the infinite series,
where a sub n would just be
one divided by n to the 4/3,
with p equal to 4/3,
which is greater than one, converges.
Okay that's going to do it for this example.
We'll take a look at another example in the next video.
I hope you found this helpful.