I am mr. Tarrou In this calculus lesson
we're ultimately going to do three
examples determining the interval of
convergence of a power series what's
power series well it says here if F is a
variable then an infinite series in the
form of the summation where n starts at
0 and goes to infinity at the a sub n
times X minus C raised to the N power is
equal to a sub 0 plus a sub 1 times X
minus C u raised to the first power plus
a sub 2 times X minus C raised to the
second power and we keep adding and so
on and so on until we get to the nth
term and then yeah we just keep adding
because this is a series you know and
goes to infinity we never stop adding
these terms and this is a power series
centered at a constant C now if I were
to come through here and take out all
these variables or this not really
variable but this constant of C and just
have it set equal to 0 then that would
be a power series excuse me centered at
0 but this is the more general form so
one thing to note here because now we're
talking about you know convergence and
divergence and we have these Sigma
notations or these summations
expressions showing up not the sort of
standard looking series that we've been
working through in our previous sections
we talked about ratio and root test and
alternate series tests and so on and so
on but you'll see here that we go when
we go from the summation notation into
our expanded series notation the
variable of n is gone that's starting at
0 and going to infinity we have a
function that is in terms of X so when
we are trying to determine the interval
of convergence what we're actually
trying to determine is when we plug in
some value of x we get a series and does
that series converge or diverge do we
get a real sum from that series if we
get a real sum that occurs right when
our series converge and so if we get an
actual sum for the series if this series
is converging well then this is going
give us write that valid well says right
here some so we are oh and by the way
you can kind of notice when you see an
expanded series like this if you have a
variable X raised to some power that's
when you identify and you can go okay
that is a power series this series
converges like I just said a second ago
if the series converges we get a valid
sum we are determining for what values
of X does that summation give us a valid
sum we are looking for the domain of all
X's such that this series right here
converges if it converges we get a valid
sum we're good to go if it diverges well
then our sum would be something like
infinity or undefined that's not what we
are looking for but that's hopefully
giving you a good understanding here an
idea of what is going on now to do these
three examples we're going to use
something that we're going to use that
ratio test we just talked about that in
the previous section because that ratio
test says that a series absolutely
converges if the absolute value of the
ratio of a sub n plus 1 over a sub n is
less than 1 if it's equal to 1 it's in
determinant if it's greater than 1 then
that series diverges so the ratio test
gives us an inequality to work off of to
determine that interval of convergence
so as I read through here what we're
looking for I'm talking about radius of
convergence I'm talking about doing that
ratio test to this series and we're
going to make sure that that ratio stays
less than 1 if the radius of convergence
is equal to 0 then the power series is
going to be said to converge only at
that constant at that C the center of
that power series if the radius of
convergence is infinity and I'm pausing
because I'm going ok I want to say these
0 and this infinity and then found the
condition here with the outcome number 3
if you will is that breeze
again of setting up the ratio test and
looking at ultimately what we're
arriving at it'll be clear as we do the
examples if the radius of convergence is
infinity then the power series converges
absolutely for all X so doesn't even
matter what X is so these two types of
solutions will be example 1 and 2 and
they're they're not quite so interesting
because there's no work in determining
what that interval of convergence is now
if we do the ratio test we set that up
and there exists a real number R that's
greater than 0 so not zero not infinity
that's where the little bit of extra
work that's where the skills that we
learn in all these previous sections
learning all these tests alternate and
you know alternating series test and
direct and limit comparative test so and
so on come into play well if there
exists a real number R greater than zero
such that the series converges
absolutely for the absolute value of X
minus C is less than R and diverges for
the absolute value of X minus C is
greater than R we're going to be
focusing really on this part here when
you get the absolute value of some
binomial alone and it's less than well
whatever your radius of convergence is
going to be we'll be able to identify
what the center of our power series is
if we can't already see it in the
expanded form and through the for the
three examples that we're going to be
doing we're going to have our series
written like in this format here and
we'll have our radius convergence now
from there we're going to find that will
be fine we'll have found that radius of
convergence and then we're going to test
whether the power series converges or
diverges at the endpoints of your
interval that's where maybe you need to
once you plug in some particular x
values into the series start looking at
it and trying to go cables this harmonic
alternating harmonic series is it a
geometric series is it something that we
need to do some kind of limit
comparative test to you need we need to
determine whether this series converges
or diverges at the endpoints of our
interval to come up with our final
answer now
this is a lot of words and writing and
hopefully kind of make sense but I think
doing these three examples is going to
clarify this up for you the first one
coming up right now so all the examples
are gonna have the same directions find
the interval of convergence of the power
series so we have our power series the
summation where n starts at 0 and goes
to infinity of 2 n factorial times X to
the 2n over n factorial and just to
expand it out for you here when let when
we let n equals 0 we're going to have 0
factorial and 0 factorial well that's
defined to be 1 and X to the 0 power is
1 and then when we let n equal 1 we have
2 factorial over 1 factorial which is 2
and then X to the 2 times 1 which is x
squared and then plus 12
with our N equals 2 we have 2 times 2 is
4 4 factorial is 24 over 2 which comes
out to be 12 and we have X to the fourth
and so on and so on
what's the interval where this series
converges ok so we're gonna do that
ratio test so by the ratio test we're
going to be looking at the limit as n
goes to infinity with this expression
here of a sub n plus 1 just like the
previous section so we have 2 to the n
plus 1 factorial X to the 2 times n plus
1 power over n plus 1 factorial over 2
to the n times X to the 2n over n
factorial so let's get all this cleaned
up until we get that that binomial that
X minus C the absolute value that
compared to our radius of convergence
we're going to take this denominator and
flip it up
we're also going to distribute this to
through the parentheses of n plus 1 and
expand that factorial hopefully with all
that practice from that section when we
first learned the ratio test you saw a
lot of this expansion and you got used
to expanding these factorials but just
to show you one more time to plot 2 n
plus 2 factorial is 2 n plus 2 count
down by 1 2 n plus 1 and then count down
by 1 again and we get to that 2 n
factorial so I'm going to just stop
there and write 2 n factorial and we can
see all these nice factors cancel out oh
and of course we have down here and plus
1 times and counting down by one until
you get to 1 so n factorial so we have
the factors of n factorial cancelling
out well that's been already rewritten
and we have two n bases of X in the
denominator and we have 2 n plus 2 bases
of X in the numerator so there's two
more X's in the numerator than in the
denominator so that is going to cancel
out all but two of them and we've got
our two n factorials cancelling out
let's clean this up a little bit and see
where we stand
okay I don't know why I took the time to
actually distribute this out for the
numerator because it's not really truly
necessary when we let n go to infinity
the only thing that's going to control
of course the limit of this first factor
is the leading term but at any rate we
have for N squared over N now
l'hopital's rule or dividing the top of
the numerator and denominator by n
whatever it takes that we can see that
the limit here of our first factor is
going to be infinity times X now that
absolute value of that has to remain
less than 1 and we're about to say here
you know infinity and that doesn't
really need to be in the absolute value
symbols now because we know it's going
to pause infinity but infinity times X
infinity times what is going to
guarantee that you have an answer that's
less than 1 well if you put any value in
for X now that doesn't even apps you
guys almost because we could plug in
anything but any value you plug in for X
that's not equal to what 0
I mean 1/10 times infinity is still
infinity so the only thing the only
value of X is going to guarantee that
this expression the absolute value of
this expression is less than 1 is if X
is equal to 0 well kind of a special
case we got here and by the way when we
see when I'm saying that X has to be 0
you might going you might be going well
if we have infinity times 0
isn't that indeterminate form when we
were studying l'hopital's rule and we
had to then rework the limit so we had
it to be infinity over infinity or 0
over and Vidur 0 over 0 some form that
allows us to apply l'hopital's rule well
no because we're saying that X you know
to make this inequality be true the only
value that X possibly can be is 0 X is a
constant of 0 it is 0 not approaching 0
this for N squared plus 6n plus 2 over n
plus 1 that is approaching
infinity now if we had some kind of end
variable over here that was approaching
something to make the second factor
approach zero well then we would have
some form of an determinant and we'd
have to you know manipulate that with
l'hopital's you know so that we could
eventually apply l'hopital's rule and
evaluate the limit but know this in this
case is not in a determinant form it is
actually equal to zero so the only time
not much an interval here
the only value of X which is going to
make this series converge is x equals
zero yeah so our radius of convergence
is zero this converges this given series
converges when x is zero and by the way
the center of our power series do we
have any kind of you know X minus a
number here is it just X is just X and
actually it might be beneficial for me
to let's go ahead and pull you know n n
starts at zero and goes to infinity so
this is never going to be negative so
really you just need the absolute value
symbol around the X and I guess you
could say you could divide both sides by
infinity and have the absolute value of
x is less than 1 over infinity which is
zero allowing me to say that that radius
of convergence is equal to zero and thus
x equals zero let's go on to our second
example right now already our second
example we're going to look for the
interval of convergence of this series
here we have n the summation of n
starting at 0 going to infinity of 4
times X to the N power over 2 n
factorial and I did a little bit of
expansion here
just to let you see that we again have
that function in terms of X we're
finding the domain of this function such
that it gives us a valid some such that
it converges so we're going to set up
that ratio test and by the ratio test we
have the limit as n goes to infinity of
a sub n plus 1 over a sub n again
instead of dividing by that fraction
we're going to bring it up as a multiple
over reciprocal and clean things up this
factor of 2 in factorial now that we've
done a little bit of expansion on that
factorial notation is going to cancel
out with 2 and factorial in the
denominator of our second factor we have
n bases of 4x here we have just one more
so this is going to cancel out and leave
us with just one factor of 4x in the
numerator and this is really kind of
simplifying very similar to the previous
one but we have a lot of ends now in our
denominator as opposed into the
numerator of our last one now if we let
n go to infinity we're gonna have 1 over
infinity which approaches zero hmm and
zero times 4x is just zero so the limit
here I was going to write it on the
second line but I don't need any extra
room to write zero the limit as n
approaches infinity is zero and of
course zero is always less than one so
since zero is always less than one and
there's not even a place for X to go
right so for all X on the interval from
negative infinity to positive infinity
so therefore
by the ratio no yeah so therefore the
given series converges absolutely for
all pets I'll be right back and for our
third and last example we have the
summation where n starts at 1 and goes
to infinity of and negative 1 to the n
plus 1 which means that we are looking
at an alternating series by the way
times X minus 3 to the N over N times 7
to the N power okay so we're looking for
the interval of convergence of this
series this power series which means
that by the ratio test we are looking at
the limit as n approaches infinity of a
sub n plus 1 over a sub n now we put
this in the notes earlier but how is it
you know how do we look at this and you
know how can we tell that we have a
power series now maybe you could expand
it a couple of terms to make sure you
know what those ends kind of disappear
and look at what your expansion looks
like because you're only you're only
going to have that variable X you have a
function in terms of X and you have an X
raised up to some power that power of
course is gonna be increasing as we let
and go from 1 to infinity so that
variable x raised to some power is your
indication that you're looking at a
power series okay as always with ratio
test we're going to multiply by the
reciprocal and clean things up lots of
cancellation so the limit as n
approaches infinity of the absolute
value of the negative 1 raised to the
now n plus 2 power because of course
one plus one times X minus three raised
to the n plus one over n plus one times
seven to the n plus one power multiply
that by the reciprocal of the
denominator we have n times 7 to the N
over negative 1 to the n plus 1 times X
minus 3 to the N power okay so we have n
7s in the numerator of our second
fraction and n plus 1 in the denominator
of our first fraction leaving you with
one factor of seven in the denominator
we have X minus three to the N and X
minus 3 to the n plus one just one more
factor in the numerator and the same
thing here - we have negative 1 to the n
plus 1 and negative 1 to the n plus 2
well it was just one more than 1 now
cleaning this up what do we got we have
the limit as n approaches infinity of
the absolute value of well and actually
let's do this let's take out all those
ends because they're going to be
positive the absolute value again is
going to take care of that factor of
negative 1 we have an N so the limit as
n approaches infinity of n over that's
this one right here in the denominator
we have a factor of n plus 1 and we can
bring that 7 out as well if we want or
we can leave it you know inside the
absolute values I think I'm going to
leave it inside the absolute value not
that 2 value of 7 is just going to be 7
so we have let me just make sure I
didn't copy something down wrong
absolute value taking care of that
negative 1 we have our factor of x minus
3 here we have a 1 over 7 the absolute
value the N is starting at 1 and going
to infinity so it's never gonna be
negative pulling that out in front of
the absolute value so I'm just giving
this a little bit of a different look
the limit as n approaches infinity is
this n over n plus one is going to come
out to be one so we have that this is
equal to the absolute value of 1 over 7
times X minus 3 the absolute value of
all of this and remember we are looking
for the domain of this function
well actually this function when it was
expanded that you know the domain of X
that makes the function converged so
that we have that valid sum so that with
the ratio test says that the absolute
value of the ratio of a sub n plus 1
over a sub n must be less than 1 for
that to absolutely converge and the
absolute value of 1/7 is 1/7 so actually
I could bring this go ahead and bring
that out and at this point like I
mentioned earlier in the notes once you
have that X minus C the absolute value
of X minus C is less then well it'll
have a quite the absolute value symbol
alone yet so I'm going to multiply both
sides by 7 and we have the absolute
value of X minus 3 I'm gonna start of
talking there could you tell and say
something that was incorrect we have 1
times 7 is equal to 7 now here's at the
point you can look at this and go hey
the center of our power series is 3 and
here we have that Capitol or that real
value that's greater than 0 we have a
radius of convergence of 7
okay so we have now a result that is
going to allow us to set up an
preliminary interval for convergence but
then we're going to test the endpoints
so let's go ahead and write this without
the absolute value signs which means
that this can be well if the absolute
value in the sevens can be like 7 or
negative 7 that it has to be in between
those values so we have negative 7 is
less than X minus 3 which is then less
than 7 solving this compound inequality
we going to you know very difficult here
add everything by 3 and we have negative
4 is less than X which is less than 10
so our interval not I can't say really
interval of convergence because we
haven't tested the endpoints yet but our
preliminary interval here is from
negative 10 or excuse me negative 4 to
10 so now we need to test the endpoints
and I'm going to try and squeeze this
down in here I'll pick a different
colors you can tell the difference of
what I'm doing test the endpoints very
easy for students to forget this part my
textbook that I teach from even
acknowledges hey don't forget to test
the endpoints so what about when X is
equal to negative 4 and I'm really
afraid I'm going to run out of room here
but we're going to be looking at the
summation where n goes from 1 to
infinity and I'm just going to plug I'm
just going to take out the X and plug in
negative 4 so negative 1 and plus 1
parenthesis negative 4 minus 3 to the N
power over n times 7 to the N now here's
at the point that no where's with some
simplification that we can go through
but here's where the point that we you
know you might find out that a lot of
these become very easy to determine
whether they converge or die
but you might need to lean on those I
you know start pulling those toolboxes
out are you gonna use maybe a p-series
test a alternating series test interval
integral test limit comparative test
direct comparative test route test ratio
test that here's where we come in and we
have to use all of those tools and know
which one is appropriate and many of
these mean you know when you're test
whether series converge or diverge most
of them have more than one way to figure
that out but what do we got going on
here we have the series negative 1 to
the N plus 1 negative 4 minus 3 of
course is negative 7 to the N and we
have n here with the 7 to the N power so
we have a negative 7 to the N and then
our denominator we have positive 7 to
the N looks like we could almost make
those things cancel out right and what's
the square root of 2 times the square
root of 3 well that's just the square
root of 6 right you can multiply inside
well what do we have here we have 2 to
the 1/2 power times 3 to the 1/2 and
that's 6 to the 1/2 so we can multiply
together 2 times 3 is 6 and then reverse
factor that 6 to be 2 times 3 within
those matching rational powers in this K
12 it because he has a rational powers
in this case we just have n but I'm just
pointing it out to validate for you here
that I can take the negative 7 to the N
power and write it like this
basically just factor it so we have
negative 1 to the N times 7 to the N
over N times 7 to the N well check this
out now if I rewrite that 7 as negative
1 I forgot the plus 1 here as negative 1
times 7 well then the 7 to the N power
over 7 to the N power these are all
factors those common factors will cancel
out we have
a multiplication of like bases well neg
you know when you multiply like bases
you get to add those exponents and n
plus n is 2 n so in the numerator we're
left with negative 1 to the 2n plus 1
over N and 2n plus 1 I think you're
probably pretty comfortable by now if
you take a number and you double it and
you add 1 you're guaranteed to have a
odd answer and negative 1 to an odd
power is always going to come out to be
negative 1 so this looked like an and
was an alternating series but when we
check the end point where X is equal to
negative 4 we actually get negative 1
times the summation or n starts at 1
goes to infinity of 1 over N so it's
negative 1 times a summation of a
harmonic series and harmonic series are
divergent so when we tested our endpoint
of x equals negative 4 at negative 4
this series diverges so that is not
going to be included in our final
interval of convergence now let's check
the 10 and I did run out of room so
hopefully this is okay I'm gonna go
ahead and erase this
so should went ahead just race it all of
it huh all right taking the next value
of 10 plugging it in what do we got we
have negative 1 to the n plus 1 times 10
not n anymore or X for that matter 10
minus 3 to the N power over n times 7 to
the N well of course 10 minus 3 is 7 and
our 7 to the N power like the previous
one 7 over n divided by 7 over to the
nth powers whose me is equal to 1 and
this time well this simplification for
when the check and the endpoint of x
equals 10 comes out to be really nice
and simplifies very quickly we have the
summation where n starts at 1 and goes
to infinity of negative 1 to the n plus
1 over N well now this negative 1 to the
n plus 1 that's going to well it's going
to go between positive and negative and
thus we have an alternating harmonic
series an alternating harmonic series
unlike the regular harmonic series
converge so we took our original power
series we set the limit up we did a
ratio test at the women up as n
approaches infinity of a sub n plus 1
over a sub n we did a lot of
simplification unlike the previous
examples the two examples we did get X
minus 3 the absolute value of x minus 3
or X minus C is less than some real r
which is greater than 0 in this case it
was 7 so our Center is 3 our radius of
convergence is 7 we went ahead and wrote
that as a compound inequality we solved
found in preliminary interval of
convergence check the endpoints negative
4 didn't work in the sense that it came
out to be that series came out to be
divergent but 10 did at x equals 10 our
series did converge so our final
interval of convergence and just 2
because I'm running out of space here
this is our interval
of convergence so again if you take
these X values starting from but not
including negative for up to ten plug
them into the function that you get as
you expand out the summation notation
because all the ends would go away this
is actually equal to a function in terms
of X plug these values of X into that
function of X and expansion of that
summation notation you will get a series
that converges this is the domain that
allows this series to converge thus
giving you a valid sum I'm mister true I
am go to your homework
you
