I am mr. Tarrou.  Do you have a series
involving nth powers?  well then I've got
the test for you and that is called the
root test this would be the last test
that we put into our tool box tool box
before moving on to Taylor polynomials
we have been looking at series in
determining whether they converge or
diverge using a recognizing that their
geometric series telescoping series we
have done the P series test the integral
test the alternate series test the limit
comparison test the direct comparison
test the ratio test and in your book
it's probably in the same section like
it is in mine but we have here with the
root test and it says let the summation
of a sub n be a series and that series
converges absolutely if the limit as n
approaches infinity of the nth root of
the absolute value of n is less than 1
and I'm only gonna do two examples to be
pretty quick video
neither example though involves like an
alternating series or has any negative
values so but if you have like a
negative 1 to the N minus 1 power or
something like that the absolute values
is going to take care of that our series
is going to diverge if the limit as n
approaches infinity of the nth root of
the absolute value of a sub n is either
greater than 1 or infinity and then
finally if you get a limit that's equal
to 1 well then you've done some work but
you have more work to do because the
test is inconclusive we have not
determined whether the series converges
or diverges maybe move on to say I don't
know maybe an alternating series test or
direct comparative test or something
like that but here we have example 1 it
is the summation where n goes starts at
1 and goes to infinity of 3 to the N
plus 2 power excuse me 3n plus 2 over 2
and minus 1 to the nth power now if that
nth power was not there we would just do
the limit as n approaches infinity of 3n
plus 2 over 2 and minus 1 and the nth
term test would fail because that that
limit is not equal to 0 but we have that
little bit extra complication there with
a power of n so the entire expression in
my series has that nth power let's see
what have
we do the route test so we're gonna be
looking at the limit as n approaches
infinity of the nth root of the absolute
value of a sub n you know I don't have
any kind of negative 1 to the nth power
but you just went in and put that in
there but that would get canceled out
anyway by the absolute value now these
this power of N and this root of n are
going to go ahead and cancel out now if
these if s power on this root are even
then the power and the even root are
going to act as an absolute value symbol
but of course we already have those here
so now we're looking at the limit as n
approaches infinity of 3 n plus 2 over 2
n minus 1 that that nth root part of
this root test is actually now that it's
been canceled we're kind of like going
to that nth term test to see if the
series converges or diverges and here
we're going to have infinity over
infinity but you can divide the
numerator and denominator or multiply
through the numerator and denominator 1
over n to get it into a format that
allows us to plug in infinity or you can
apply l'hopital's rule and take the
derivative of the numerator and the
derivative of the denominator
individually and however you approach
this problem of course this is a pretty
simple simple pretty simple limit to
analyze it's gonna come out to be three
halves now three halves it is greater
than one and so therefore by the ratio
test
the given series diverges and at least
out of the textbook I'm teaching from
and the problems I'm going to assign all
these are problems that these series
that you can test using the route tests
do have a pretty simple or short
simplification process that's why this
video is only a couple of examples long
but our second example is going to take
a little bit more work to find that
limit as n to approaches infinity and it
doesn't involve any negative values
requiring the absolute value symbol but
it is a more interesting question than
this one and for a second and last
example we have the summation where n
starts at 1 and goes to infinity of n
over 2 to the N and unlike our first
example that had a power of n in the
numerator and the denominator which
admittedly was a pretty simple example
we just have that power of N in the
denominator so maybe the root test
wouldn't be your first you know go to
test to try and analyze whether the
series converges or diverges and to be
honest I haven't tried to test you know
use the other all of the other tests on
the series we're just going to see what
happens when we try to apply the root
test because it's nice and when it when
it works it's just nice and easy there's
a limit of the loss to it the limit as n
approaches infinity of the nth root of
the absolute value of a sub n does that
come out to be less than 1 does the
series converge by the root test well n
is starting at 1 and going to infinity
and this is of course a series which
means that n will never be negative so
these absolute value symbols you know
the absolute value of this expression is
simply this expression n over 2 to the N
now we're going to rewrite this
basically this radical with a and a
fractional or rational exponent just
showing all of the steps because
honestly you know I need them okay so
you know basically just a power of one
here and through we got one over n
applying that power to both the
numerator and denominator using the
power to power rule and then of course n
times 1 over N is just going to be 1 so
we're taking well we have basically a
factor of 1 in the numerator so we have
1/2 times n to the 1 over N power so we
can bring that 1/2 out front and let's
see what we have well as we let n
approach infinity we've got n is
approaching infinity so it's approaching
infinity and our our exponents going to
be 1 over infinity which is 0 0 or
infinity to the 0 power is one of the
forms have been determinate we learned
and of just a few seconds ago when we
were talking about l'hopital's rule so
we need to kind of take this out here to
the side and say do some scratch work we
have y is equal to the limit as n
approaches infinity of n to the 1 over n
let's try and manipulate this so we can
get it into a form that will allow us to
apply l'hopital's rule because we can't
do it right now we need like a 0 over 0
or infinity over infinity right to allow
us to apply that l'hopital's rule
we got a variable base and a variable
exponent okay we need to we need to move
that exponent so we need logarithms to
allow us to do that so we're gonna take
the natural log of both sides
now that's going to allow us to bring
out this exponent of 1 over N out in
front of the natural log function now
let's see we have right now if we let n
approach infinity now we have 0 times
infinity so we've manipulated this you
know this expression basically into just
another form of indeterminate that still
does not allow us to use l'hopital's
rule
we need that 0 over 0 or that infinity
over infinity we need some kind of
fractional format so let's turn this
multiplication of 1 over n into a
division show that you know multiplying
by 1 over N is effectively dividing by n
right ah so now when we try and find the
limit we have a still another form of a
determinant but we have infinity over
infinity which is a indeterminate form
that allows us to use law at all so we
have why using and I'm just going to do
that kind of like a little owl
apostrophe H to let people know that I'm
doing l'hopital's rule and we have the
natural log of y is equal to the limit
as n approaches infinity the derivative
of the natural log of n of course is 1
over N the derivative of n is equal to 1
and now as we let n approach infinity we
come out to be equal to 0 ok so we have
1/2 well let's keep actuate on hello
come back here we still have to take
care of this natural log of Y the way it
going on so we have the natural log of Y
is equal to 0 ok
why ok that's interesting the natural
log of Y is equal to 0 I kind of stopped
this problem so I could work or that
right that lemon notation which I set
equal to Y and you know what Y is equal
to so we're gonna make both sides an
exponent of
and we have y is equal to e to the first
witches 1 so that means that 1/2 times
the limit as n approaches infinity of n
to the 1 over N power is equal to 1/2
times 1 which is equal to 1/2 now 1/2 is
less than 1 so therefore by the ratio
test our series does what converges
absolutely even though that was never
going to be negative in the first place
I miss true that's the root test I'm
looking up here like the name still
there anyway let's try this again
I miss true hey go to your homework
you
