Bam Mr. Tarrou.  In this calculus lesson we
are going to put another tool in our
toolbox and that is called a ratio test
we've been determining whether series
converge or diverge and we have looked
at geometric series telescoping series
we've used the P series test the
integral test the alternating series
test the limit comparative test the
let's see direct comparative test and
now we're going to be looking at the
ratio test we need all these tools it's
a difficult chapter and these these
series can vary widely we need all these
tools so we're going to be looking at a
series that involves factorials or
Exponential's series that may converge
very rapidly and in that case the ratio
test may be very useful this test is
going to be used for finding intervals
of convergence in a couple of sections
so we're not going to be just using this
test you know right now and kind of
forget about it's going to come right
back around be really useful for us in
the near future the ratio test will be
inconclusive by the way for any p series
so what does this test say well the
ratio test says let a series or the sum
of a sub n be a series with non zero
terms so all the a sub n terms must be
nonzero
well that series converges absolutely
that step above just regular convergence
if the limit as n approaches infinity of
the absolute value of a sub n plus 1
over a sub n is less than 1 now this
series is going to diverge if the limit
as n approaches infinity of that
absolute value of a sub n plus 1 over a
sub n is greater than 1 or if your limit
comes out to be infinity now if that
limit ends up coming out to be equal to
1
well then simply you've done some work
and unfortunately you have more work to
do this means the test is inconclusive
it doesn't mean that the series either
converges or diverges it just means that
you'll have to try some other tests in
our toolbox to take a different approach
at it like probably pretty decent chance
that maybe you could work with the
alternating series test to see if the
series converges where this ratio test
comes out to be inconclusive now we're
going to be doing four examples in this
video
they're gonna go from easy to hard you
might even know the answer to the first
example before I even start but that
just simply means that you're noticing
that the series we're working with is a
kind of a special type of series if you
will and we could approach or arrive to
the answer much quickly but then you'll
see that the ratio test gives us that
same conclusion and you know maybe it
just helps you understand that it works
now there's I'm not gonna be working
with proof of this ratio test you can
find that in the appendix of your book
or online if you like we're gonna follow
this video up with a route test that
one's only going to have two examples in
it and then in the sections coming up
after that we're going to be talking
about Taylor polynomials Taylor series
determining you know if you have what
are the intervals for convergence for
power series so we got a lot more coming
up even though we're near the end
nearing the end of this chapter let's
get to that first example right now for
example number one and the directions
are all the same determine if the given
series converges or diverges well we
have the series where n starts at one
and goes to infinity of 1 over 3 to the
N power now we're going to do this we're
going to analyze whether this converges
or diverges with this new ratio test
that we have so with our ratio test yes
indeed it's true that one of course
let's remember that if you're trying to
determine if the series converges or
diverges it's always a nice idea to do
the nth term test I forgot to mention
that at the beginning of the video just
make sure that the limit as n approaches
infinity of a sub n let's make sure that
that actually equals a zero because if
it doesn't equal zero well then it's
automatically going to diverge and all
this other work is kind of pointless now
our first condition is that you know all
of our a sub n terms are non zero and
Abid you know three to any power 1/3
dinning power this is never going to be
equal to zero
so we've got that condition out of the
way we're going to be testing or looking
at the limit as n approaches infinity
of the absolute value of a sub n plus
one I mean just take out the N and
replace it with n plus one so there's
our a sub n plus 1 term the next term in
our sequence and if you let that
sequence go on forever we have a series
but at any rate we have one for three to
the N of course I'd be adding all the
terms as well but we have our limit here
as n approaches infinity of a sub n plus
1 over a sub n we're going to clean this
up a little bit of course we're dividing
my fraction so we're going to flip that
bottom one up I I like to keep the
absolute value symbols through all of
these steps of my work until I get to
the very end just as good measure you
know really a positive 3 raised to any
power R are here in numerator and
denominator or one neither one of these
factors ever going to be negative and
thus this term is never going to be
negative it's a positive times a
positive so the absolute value symbol is
here really for this particular problem
are not necessary they're redundant but
if there's ever any doubt just leave
them in there because you're guaranteed
to be right if you if you drop the
absolute value symbols a little bit too
soon and by chance your expression your
can possibly possibly be negative
somewhere you're just setting yourself
up to lose some points on a test or
homework assignment or such that
otherwise you wouldn't have left missed
if you just let the absolute value
symbols in there now we have 3 to the N
power and we have 3 to the n plus 1 just
making sure that you understand here
that that just simply means exponents
tell you how many bases you have and n
plus 1 means you have one more 3 in this
base here basically as you have 3 to the
N or you can write this as 3 to the N
times simply 3 and when you multiply
like bases you add the exponents you're
gonna be doing so much of this this
cancellation that you probably are
thinking why am i bothering taking
taking time to do this or if you're
strong calculus student you don't need
to have that explanation but some of us
do so we have 3 to the N in the
numerator we have 3 to the n plus 1 one
more 3 in the denominator of course all
of these are going to cancel out leaving
you with
of those factors of three in the
denominator I could have crossed out
those three to the N powers there and
you see that is left now we have the
limit as n approaches infinity of well
it's just simply one third and there's
not even a place to you know let n go to
infinity in this expression so our
answer is simply 1/3 now remember over
here when I was defining the ratio test
if the limit as n goes to infinity of a
sub n plus absolute value absolute value
of a sub n plus 1 over a sub n if that
limit is less than 1 well therefore I
the ratio test the given series I like
to just write it out converges now as
soon as you saw this example up on the
board and me saying hey I want to do the
ratio test to show you that this or
determine I already knew that I was
going to say show you that it converges
because already new converges I made the
example you might have been saying why
are you even bothering doing all of this
because another way of writing this
series is to just say that's one third
to the N power I mean yes it says 1 over
3 to the N but 1 to any power is still
going to just simply be 1 well look at
that that is excuse me a geometric
series with an R value our R value is
1/3 and that is less than 1 so we have a
geometric series within our value that's
less than 1 therefore it converges or we
can do with the ratio test now with the
ratio test you can actually say it
converges absolutely or absolutely
converges because this is a test that
will determine
the convergence so that is the end of
our first example not very exciting
because we could have seen that it was a
geometric series and knew from that
knowledge that we have that it converges
anyways so let's look at another example
a little bit more interesting so here
for our second example we have a series
which is got negative 1 to the N power
which means as n goes from 0 to 1 to 2
to 3 that we're gonna be you know
alternating back back and forth between
positive and negative numbers which may
be but that means that this would be a
good series to apply the alternating
series test but I also said with the
ratio test if you have a series that
involves Exponential's and factorials
look at their a fractional exponent and
factorials so probably the ratio test is
going to be very very helpful to us and
be my first choice to determining
whether the series converges or diverges
so let's make sure that indeed all of
our a sub n terms are nonzero yep never
going to be equal to zero so that
condition is met we're going to set up
the ratio test and look at the limit as
n approaches infinity of a sub n plus 1
so we're going to take this expression
and anywhere there is an N I'm going to
replace it with well a sub n plus or n
plus 1 so we've got negative 1 to the n
plus 1 power times 3 to the n plus 1
over now here we have n plus 1 plus 1
factorial which of course is going to be
n plus 2 factorial now that's our a sub
n plus 1 term over a sub n just copy
down we have here now these absolute
value symbols are once again going to
take care of these negative factors but
again I don't want to drop them I'm
gonna leave these absolute value symbols
through all of the work even if I could
say well we're going to drop these
absolute values
as we cancel out these factors which is
negative but remember I said in a couple
of sections we're going to be using the
ratio test to determine intervals of
convergence and that absolute value
symbol at the end is going to help you
help remind you of exactly how to set up
that interval so just leave them in
there so we're going to bring up the
denominator and now we have to simplify
this now I apologize if I spend a little
bit too much time going over the
simplification process I don't know if
you're an extremely strong calculus
student and you just miss this lesson or
you're like do you like to learn online
you're doing some independent study if
your student really struggles but I'm
gonna take a little bit of time to talk
about the simplification here these of
course these exponents tell you how many
of those bases you have and we have n
threes and denominator and in the
numerator up here we have n plus just
simply one more or again you can write
that as 3 to the N times 3 to the first
and those 3 to the ends are going to
cancel out there's one more 3 in the
numerator than the denominator so this
is gonna cancel out leaving you with
just one factor of 3 in the numerator
same thing with our bases of negative 1
so there's n of them here and there's n
plus 1 in the numerator just leave me
with that one negative 1 in the
numerator
now our factorials hopefully by now you
have gotten very used to expanding these
and seeing how they cancel out my
precalculus students even at the end of
the year this year are struggling with
this a lot so I'm going to show you that
this n plus 2 factorial course
factorials mean you can't count out
words count down by one until you get to
one so this is going to be n plus two
now reduce that by one so 2 minus 1 of
course is 1 so on and so on just keep
multiplying back until we get to 1 so in
fact we're like 5 factorial 5 4 3 2 1
stop multiply them all together we see
here n plus 1 factorial is going to be n
plus 1 reduce 1 n
reduce one and minus one and reduce one
and reduce one until you get back to one
so I can expand this n plus 1 factorial
out but I'm hoping that your that you
see here that n plus 1 factorial is n
plus 1 times n and so on and so on keep
counting back by one until you get to
one so this n plus 1 factorial cancels
out with this expanded notation of n
plus 1 factorial and we're just left
with n plus 2 in the denominator so now
we've got this okay well as we let n
approach infinity we're going to have
negative 3 over basically infinity which
means that this limit is equal to 0 now
0 is less than 1 so therefore by the
ratio test actually instead of just
saying converges absolutely converges
all right third example coming up right
now now we're third example well we have
a series that involves Exponential's and
we dealt with Exponential's we handle
them a little bit in the previous
example so this is going to be very
similar I don't worry about the you know
negative 1/2 an N or n plus 1 power
there's no factorials so maybe in a way
this is a little bit simpler but let's
walk through it anyway so we're going to
be taking a look at the ratio test again
all of my a sub n terms are going to be
non-negative so set up that limit as n
approaches infinity of 7 to the n plus 1
3 to the n plus 1 plus 1 over some to
the N over 3 to the n plus 1
the simplification process of all of
these ratio tests you're gonna be seen
in your book for this section they're
gonna look very very similar I'm gonna
be flipping this denominator up instead
of
biting by that fraction multiplied by
the reciprocal now getting cleaning this
up now I want to point out a mistake
hopefully especially now we're in count
two you're not making this mistake but
you know I teach a lot of precalculus
students and calc one make sure you
don't cancel terms this is 3n plus 1 and
this is 3 n plus 1 plus 1 so this
numerator and this denominator have two
terms and them so don't be tempted to
cancel this 3 to the N power with this 3
to the n plus one the only thing that's
going to cancel here is that we have a
factor or many factors but we have a
factor of 7 to the N power and up here
we have 7 to the n plus 1 again there's
one more base of 7 in the numerator than
there is in this denominator so we're
gonna cancel n of those out leaving us
with just 1 and we have this now our
factor of 7 is positive we don't need
that absolute value symbol to take care
of that I mean again I leave those
absolute value symbols in there to
remind me that they should be there not
lose any points when maybe I've actually
just sort of ignored them when they
shouldn't have been and also for helping
me set up that interval convergence in a
couple of sections I just like to leave
that notation in there but we can take
this 7 and move it out front
ok so we're looking at the limit as n
approaches infinity of 3 to the n plus 1
over 3 to the n plus 1 plus 1 now what's
that going to come out to well maybe you
know the answer but we can show this
limit notation this limit work look for
that largest denominator denominator
we're gonna ultimately the n plus 1
power same with the denominator so of
course anything divided by itself is one
certain change the look of this
expression but not its value
so we have three to the N over 3 to the
n plus one is going to leave us with 1/3
because those n three is going to cancel
with those n plus one threes but leave
us with one of those bases plus 1 over 3
to the n plus 1 over now as I sort of
multiply this faction and are basically
effectively divided by 3 to the n plus 1
we're going to get anything divided by
itself as 1 and then plus 1 over 3 to
the n plus 1 ok double check my work
here great now the limit as n goes to
infinity this is going to approach zero
so we have one-third plus zero we have
the 1 in the denominator we have another
term approaching zero we have
parentheses that are too big and then
ultimately you can see that were going
to end up with a limit which is equal to
seven thirds and seven thirds is greater
than one well that's certainly you know
not inconclusive our limit is greater
than one so therefore by the ratio test
make sure you tell your teacher and your
and your test graders the test that
you're using to analyze these series by
the way
to get in series the one which a problem
divert its our last example and we have
rather a scary one we have the series
here the summation where n starts at 0
goes to infinity of n factorial over 2
times 4 times 6 dot dot and we're just
gonna keep multiplying all the way up to
our nth factor of 2 n so basically
multiplying a bunch of even numbers
together that looks quite a bit
different than my previous three
examples and probably most of the
problems you're gonna have in your
homework so it's kind of scary but we do
have a factorial in our series so we're
gonna go ahead and try setting up for
our ratio test and see what happens now
for the ratio test and you know all of
our a sub n terms must be non zeros and
this is never gonna equal zero so
looking at the ratio test we're gonna be
looking at the limit as n approaches
infinity once again of our a sub n plus
1 and for our previous three examples
I've just taken out the N and plugged in
and plus one but if I do that here you
might not see how the cancellation works
so since we're expanding the
multiplication of all these even numbers
I'm just gonna add on from the nth
factor the nth plus one factor and that
will help us see how things cancel away
quite a bit easier so here's our a sub n
plus 1 over a sub n begin the
simplification of all these ratio test
problems is very similar we're going to
multiply by the reciprocal of the
denominator we're just going to
distribute that 2 through there
okay and yeah a lot of things are going
to cancel out here we have our n plus
one factorial is n plus one times n
times n minus 1 times n minus 2 taught
down to 3 to 1 so instead of writing
that expansion of really just n
factorial and I've already done that
earlier in our examples I'm just going
to write n plus 1 times n factorial and
that's a little bit of an expansion of
the numerator and we can see that the n
factorials are going to cancel out
leaving us with just the factor of n
plus 1 and by not just taking this n out
and plug in an n plus 1 we can see by
adding the N plus 1 factor that 2 4 6
all the way up to 2n from the
denominator here is going to cancel out
with all of these factors in the
numerator over here and we get the limit
we get a small piece of talk here the
limit as n approaches infinity of just
simply n plus 1 over 2n plus 2 and
however you look at it factor out a 2
from these two terms using l'hopital's
rule simplify this like we did in our
previous example example the limit here
is going to be simply 1/2 which is less
than 1 so since 1/2 is less than 1 or it
is therefore by the ratio test the given
series I'm just gonna write the given
series
absolutely convergence this is the end
of my last example I'm mr. Tarrou BAM go
to your homework
you
