We're going to determine
if the given infintie series converges or diverges.
We also want to state the test used.
The first thing we should recognize
about the given infinite series is that
we have negative one, raised to the power of n,
which means, this is an alternating series.
Let's start by checking to see
if the nth term approaches zero,
which is the nth term divergent test.
This test tells us that
if the limit as n approaches infinity of a sub n,
doesn't equal zero, we can quickly conclude
that the infinite series is divergent,
and when considering an alternating series,
a sub n is the non-alternating part,
so for our infinite series,
a sub n will be four divided by n.
So, we'll find the limit as n approaches infinity
of four divided by n.
Well, this limit is pretty straightforward.
Notice how, for this fraction, the numerator is a constant,
and the denominator would be approaching infinity,
and therefore, this should be approaching zero,
and therefore, this limit is equal to zero.
So, when this limit is equal to zero,
the nth term divergent test doesn't tell us anything.
If this doesn't equal zero,
we can conclude that the series is divergent,
but if it's equal to zero, we still don't know
if it's convergent or divergent, which means,
we'll have to apply another test.
So, let's go ahead and say here
that the nth term divergent test is inconclusive,
which means, we'll have to apply another test
to determine whether this converges or diverges,
and because we have an alternating series,
the most obvious test will be the Alternating Series Test,
which we'll now apply.
The Alternating Series Test tells us
an infinite alternating series converges
if these two requirements are met.
First, zero is less than a sub n plus one,
less than or equal to a sub n, and,
the limit as n approaches infinity of a sub n equals zero.
Well, we saw from the nth term divergent test,
we've already satisfied this requirement,
so now, we just need to make sure that
this competent equality is also true.
Well again, a sub n
is equal to four divided by n,
and therefore, a sub n plus one
is equal to four, divided by n plus one.
Well first, notice that n is always going to be positive,
and therefore, both fractions will
always be greater than zero,
so now, we just need to make sure that a sub n plus one
is always less than or equal to a sub n.
Well, if we take a look at these two fractions,
notice how the numerators are the same, or constant,
but the denominators are different.
This denominator is always going to be one larger
than this denominator.
When we're comparing two fractions,
if the numerators are the same,
the fraction with the larger denominator
is always going to be less than
the fraction with the smaller denominator.
As an example, let's consider when n is, let's say, five.
If n is equal to five, a sub n plus one would be
four over six, or 4/6,
and a sub n would just be 4/5.
Well, 4/6 is less than 4/5, and therefore,
this pattern will continue for every value of n,
so a sub n plus one is always less than or equal to a sub n,
so we've met the requirements for this test,
and therefore, the alternating series converges.
So, by the alternating series test,
since zero is less than a sub n plus one,
which is four divided by n plus one,
which is less than or equal to a sub n,
which is four divided by n, and
the limit, which we found
when we applied the nth term divergent test,
as n approached infinity of a sub n, equals zero,
the original alternating series converges.
However, we also need to check
for conditional, or absolute convergence,
because we have an alternating series.
To check for this, we check for this
if the infinite series
of the absolute value of a sub n converges,
then the original alternating series converges absolutely.
If this series diverges,
then we have conditional convergence.
Remember, a sub n was equal to four divided by n,
and since n starts at one,
the absolute value of four divided by n
would still just be four divided by n.
So, we want to determine if
the infinite series,
where a sub n is four divided by n,
converges or diverges.
To help us do this, we could factor out this four,
and write this as four times the infinite series,
where a sub n is just one divided by n.
Now the reason this would be helpful
is that this infinite series here,
should remind us of the p-series test,
where p is equal to positive one.
So, for a quick review,
the p-series test tells us that this infinite series
converges if p is greater than one,
or diverges if p is greater than zero,
and less than or equal to one.
And therefore, this series will diverge
by the p-series test, with p equal to one,
and therefore, the original alternating series
only converges conditionally.
So, we'll say, since
four times the infinite series,
where a sub n is one divided by n to the first diverges,
by the p-series test, with p equal to one,
the original alternating series coverges
conditionally, not absolutely.
So, it is important to remember
that when we have an alternating series,
we do have to check for conditional or absolute convergence.
I hope you found this helpful.