We want to use the alternating series test
to determine if the infinite series converges or diverges.
To apply the alternating series test, we're given
that a sub n, the non-alternating part
of the series is greater than zero,
and 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 that a sub n plus one is less than or equal to a sub n.
So the terms must be non-increasing,
but actually the first two or three terms are increasing,
it doesn't really matter as long as the terms
are non-increasing long term.
So looking at our series, notice how
the non-alternating part a sub n
would be equal to n to the second
divided by the square root
of n to the sixth plus four.
Notice how here a sub n is greater than zero
for all values of n.
Now we'll determine the limit as n approaches
infinity of a sub n,
which would be a limit as n approaches
infinity of n squared
divided by the square root of n to the sixth plus four.
Now there's a shortcut for determining this limit.
If we look at the degree of the numerator and denominator,
If the degree of the numerator is higher
than the degree of the denominator,
the limit approaches positive infinity.
If the degrees are equal, the limit is equal
to the ratio of the leading coefficients.
And if the degree of the denominator is larger,
then the limit is equal to zero.
Notice how the numerator is degree two,
now looking at the denominator,
the plus four is irrelevant as n approaches infinity.
So we can think of the denominator as just
the square root of n to the sixth,
which would simplify to n to the third.
No opposite value is needed
because notice how n is always positive.
So the numerator is degree two,
the denominator we can think of as degree three,
and because the degree of the denominator is higher
than the degree of the numerator,
the denominators are increasing faster
and this limit is equal to zero.
But if we do want to show some work,
we would divide the numerator and denominator
by the highest power of n in the denominator
which would be n to the sixth underneath the square root.
So looking at the denominator,
again if we want to show work,
we can write this as the square root of n to the sixth
divided by n to the sixth plus four
divided by n to the sixth.
Now again, because n to the sixth is under the square root,
it's equal to n to the third outside the square root
so we divide the numerator by n to the third.
So we'd have n to the second divided by n to the third.
And now if we simplify, we'd have the limit
as n approaches infinity of
the numerator would be one divided by n,
the denominator would be the square root
of, this would be one plus four
divided by n to the sixth.
So now notice as n approaches infinity,
the numerator is approaching zero,
the denominator is approaching the square of root
of one over one, because this fraction here approaches zero
as it approaches infinity.
And therefore, this approaches zero over one,
and therefore the limit is equal to zero.
So notice how we've shown that a sub n is greater than zero,
we've now shown the limit as n approaches infinity
of a sub n equals zero.
So now we're left to show that a sub n plus one
is less than or equal to a sub n.
So we want to show that a sub n plus one
is less than or equal to a sub n,
at least after the first several terms.
So to save some time, I've already used
this formula here for a sub n
to generate the first four terms
which we see here.
Notice how from the first to second term,
the terms are actually increasing,
but from the second term on, the terms are decreasing.
And therefore, we can say that a sub n plus one
is less than or equal to a sub n
through n greater or equal to two,
and therefore the alternating series
does converge by the alternating series test.
So let's go ahead and summarize this.
We'll say by the alternating series test, since,
let's go ahead and say that a sub n is greater than zero,
and 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,
the given alternating series converges.
I hope you found this helpful.