Hello everyone.
Welcome to my class.
On today's class, I'm going to
cover absolute
convergence of a series,
ratio test of series,
and root test of series.
Let me start the
class right away.
So far, we have studied
in preceding sections,
convergent tests for series
for two special types of series.
Positive series, series
whose terms are positive,
and alternating series,
series whose terms
alternate the sign,
positive negative,
positive negative, so on.
Now in this section we explored
the convergence of
a general series,
and the series containing both
plus and minus
terms in any order.
For example, like given series,
summation n runs from one
to infinite cosine n over
n square has first term positive,
second term negative,
third term negative,
forth term negative, then
fifth term positive and so on.
First definition of
absolute convergence,
will so go by AC.
Our series n is called
absolute convergence if
the series of absolute
value turns convergence.
Note, summations n doesn't
want to influence
absolute a_n equals two,
absolute a_1 plus absolute
a_2 plus absolute
a_3 and plus so on.
Conditionally convergence,
in short we write CC.
A series n is called
conditionally convergence if
the series converges but
diverges absolutely.
Great. Absolute Convergence Test,
in short we write ACT.
ACT means absolute
convergence tests.
If the series
converges absolutely,
then the series converges.
Absolute convergence
series is convergent.
More precisely, if the series
afresh value converges,
if and only if both the series of
positive terms and series
of negative term converges.
In other words, if the
series n is divergent,
then the series of less
value is divergent.
Hope it makes sense to
you. Quick example.
Example 1, determine
whether the given series,
the series I mentioned
in our example before,
summation n runs from
one to infinite,
cosine n over n squared
converges or diverges.
Consider the series
of absolute value,
and here absolute a_n equals
the absolute cosine
n over n square,
and we know that value
of cosine n is at
most one over n
squared is positive.
We don't need the
absolute value here.
Since the series in
the right hand side
summation and runs
through one to infinite of one
over n squared is convergence,
bringing a p series,
where p equals to 2, that
is greater than one.
It follows by the comparison
test that the series
of cosine n over n squared and
runs from one to
infinite, converges.
This converge is absolutely
then consequently
by ACT absolute comparison test,
the series without this
value also converges.
Let's move on. Example 2.
Show that the alternating
harmonic series
is conditionally convergent.
Listen, we know by
alternate series test,
the series converges.
Let us say here. N equals to
one over n which is positive,
the second limit, limit
n goes to infinity of n,
is limit n goes to infinity
of one over n equals to zero.
Also we need to show
n is decreasing,
also n plus one means n plus one
term of this series is
one over n plus one,
which is less than or
equals one over n,
this is two, this is a_n,
so that means a_n is decreasing.
That means by
alternating series test
the series is, converges.
But we want to show the series is
conditional convergent
and that means,
we need to show
absolutely diverges.
Let's consider the
absolute value of
the given series, the
absolute value here.
Then next summation as it is,
the next n value, text
is negative one to the power
n plus one becomes positive,
so only one over n left.
Now I don't need
to put this value,
so this is now just
series one over n. We
know the series one over n
is harmonic series
which diverges,
hence the series
diverges absolutely.
Consequently the given series
is conditionally convergent.
Awesome. Let's move on.
I put some other
discussing aside also,
which is also talks about
the alternating
harmonic series here
by taking positive terms
and by taking negative term
then both series diverges.
Awesome. Let's move on.
Example 3, determine whether
the given series is
absolutely converges,
or condition converges,
or diverges.
The given series is, part a,
summations n runs from one to
infinite of negative one to
the power n times n squared
over n cubed plus 4.
This is an alternating series.
Let's pick in the solutions,
by alternating series, test
this series converges.
Since n equals to n squared
over n cubed plus 4,
factor the n cubed from the
denominator then we have n
squared over n cubed times
1 plus 4 over n cubed,
then cancel the n squared from
numerator to n square
in the denominator,
so we have only one in
left in denominator.
Then one over n,
1 plus 4 n cubed,
and take the limit as
n goes to infinite,
then the denominator
goes to infinite,
that means this
fraction goes to zero.
Next, for a decreasing,
let's use underlying
function tests.
Let f x equals to the
x is squared over x
squared plus 4 times n to x.
Take the derivative,
apply the quotient rule,
the quotient rule says,
derivative of f equals
to denominator times
derivative of numerator,
minus numerator times
the derivative of
denominator over
denominator is squared.
Simplify, this gives us x times
8 minus x cubed over
this scrapped on.
By the way, the denominator
is always positive,
so the numerator is negative
when x is greater than two.
Now we showed our
a_n is decreasing,
for n is bigger than two.
By alternating series test
and given series converges.
For absolute convergence,
let's consider
the series of absolute value,
and with this is same.
Now without this value,
no plus-minus sign here
simply summation n runs
from one to infinite,
n squared over n cubed plus
4, issues convergent test.
Let's replace this four by
n cube then this n
squared over n cubed
plus four is greater
than or equal to n
squared over n cubed plus n q
four n is greater than two.
Now we have this inequality.
The right-hand side is
n squared over two N q.
You can cancel n
square and n square,
this is simply one
over two times n,
and we know one over n,
the series of one over
n is [inaudible] series
which diverges multiplied by
constant also diverges that
means the smallest
series diverges,
means the bigger
series diverges hence,
this series absolutely diverges
that means r given series is
conditionally convergent.
Let's move on to part B.
The series given
series is summation n,
runs from one to infinite
of negative one to
the power n minus 1 times 1 over
1.1 to the power n. Let's
begin the solutions here.
Let's consider the series of
absolute value then next step
is just a summation and runs from
one to infinite of 1 over 1.1 to
the power n and this
series converges.
Being a geometric series
with common ratio R is1 over 1.1,
which is less than one.
Hence, the series
absolutely converges.
The series converges
also implies ACT.
The given series
converges. Let's move on.
There are two useful tests
for testing a series for
absolute convergence.
RATFACE, ratio test for
absolute convergence,
another one is ROOTFACE,
root test for
absolute convergence.
These tests are based on
convergent geometric series.
The root test is stronger
than the ratio test however,
the ratio test has
many application
widely used in power
series in next sections.
Let's start first ratio test.
Look at the limit of the
ratio of this fraction,
n plus one over n,
absolute value,
is effective if terms have
factorials and exponentials.
The formula is suppose the
limit n goes to infinite,
the absolute value of the ratio
n plus one over a n equals to
r. Then if r is in between
0 and 1 excluding one,
then the series is
absolutely convergent.
If the r is greater than one,
strictly greater than one,
then the series is divergent.
If r is equals to one,
then the ratio test is
inconclusive, no conclusion.
Note ratio test is inconclusive
for p-series. Quick example.
The series n runs from
one to infinite of n
factorial over 25 to the power
n diverges by ratio test,
since limit n goes to infinite
of the absolute value of n
plus one term over n term equals
to limit and goes to infinite.
Here's the n plus one term is
put n plus one here so n plus one
factorial over 25 to the power n
plus one and divide by nth term,
which is n factorial over
25 to the n. Flip it here,
then 25 to the n,
25 is the n cancels.
When your 25 left
in the denominator,
in the numerator we can
rewrite n plus one
factorial equals
to n plus one times n factorial,
then n factorial cancel out.
Only n plus one left
and take limit as n
goes to infinite,
this fraction goes to infinite,
which is greater than one.
That means the
series is divergent.
Let's talk about root test.
The root test look at the
limit of the nth root of
the nth term is effective
if terms have nth power.
Formula. Suppose limit n goes
to infinite of the nth root
of s root a n is R
then if r is between 0
and 1 excluding one,
then the series a n is
absolutely convergent.
Two. If r is still
bigger than one,
then the series is divergent.
If r is equal to one,
then the root test
is inconclusive.
If the root test is inconclusive,
do not try the ratio test
because it will also
fail and vice versa.
A quick example for root test
the series summation and
runs from two to infinite
of negative 2n over n
plus one outside power 7
n diverges by root test,
since let's take the limit
of root of nth term.
N goes to infinite,
nth root of absolute n
equals to limit n
goes to infinite.
Then has the nth root then
for two power 7 n so n
cancels so 2 power 7
times n power 7 we
got this n cancel out
in the nth root n
plus one power seven.
Then this equals to take
the constants outside the
limit then the next step,
you have two power seven times
limit n goes to infinite
of n to the power seven over
n plus one power seven.
Then pull n outside so
it becomes m to the
power seven from
denominator then n to the power
seven cancels numerator
and denominator.
The limit n goes to infinite.
The denominator here goes
to one that means the
fraction goes to one,
so we have only two
to the power seven,
which is bigger than one.
That means by root test,
when r is bigger than one,
then given series is divergent.
Some useful facts.
Limit n goes to infinite
of nth root of n is one.
We use L'Hopital's to verify
these facts limit n goes
to infinite nth root
of n factor turns
out to be infinite
limit n goes to infinite of
the nth root of n to the power n
turned out to be infinite,
limit n goes to
infinite nth root L and
N turned out to be one.
Limit n goes to infinite nth
root of n to the power of p,
turned out to be one.
Limit n goes to infinite,
nth root of p to the power n
turned out to be p for
any positive number.
Limit n goes to
infinite of 1 plus 1
over n outside n power
turns out to be e.
Two point seven one eight
two dot, dot, dot, great.
Now, let's do some example.
Example 4, determine whether
the given series is
absolutely convergent,
conditionally convergent,
or divergent.
Part a, given series is
summation and runs from 1 to
infinite and over 5 to
the power n. So here,
a_n equals to n,
over 5 to the power n,
and then a_n plus 1,
means n, plus 1th
term is just add 1,
n plus 1 over 5 to
the power n, plus 1.
Then use the ratio test,
so let's look the limit
after ratio of the a_n,
plus 1th term over a_nth term,
and this equals to limit n,
goes to infinite absolute value.
This is a_n plus 1 over a_n,
and you flip it, which is
here, in the next step.
So we can cancel the 5_n,
so only 5 left n,
plus 1 over n, as it is.
Then put one-fifth outside
in the limit of n,
plus 1 over n, turns out to be 1.
So we have one-fifth outside
which is less than 1 hence,
by ratio test the series
is absolutely convergent.
Awesome, let's move on.
Now, let's do part b,
let's compute this sum,
so the sum and runs from 1 to
infinite n, factorial over n,
to the power n. So here a_n,
is n, factorial over
n, to the power n,
and a_n, plus 1 is,
n plus 1 factorial over n,
plus 1 power n, plus 1.
If you remember, we
did this example in
our preceding section also
where you use comparison test.
Now here, let's use
ratio test, great.
Next step, limit n,
goes to infinite,
the absolute value of
the ratio of the a_n,
plus 1th term over a_nth term.
So this equals 2.
Limit n, goes to infinite,
absolute value then these
terms over these terms,
that means you just
flip this value here,
which is here, then here.
The next step, copy
the limit as it is,
we can rewrite n,
plus 1 factorial equals to n,
plus 1 times n,
factorial then cancel the n,
factorial so we only
have n, plus 1 left.
Then from down here,
we can write this is n,
plus 1 power n, times n,
plus 1, then n,
plus 1, 1, n plus 1 cancels.
So we have in the next step here,
the simplifications here,
hope this makes sense to you all.
Then we have left
this one limit n,
goes to infinite, n to
the power n, over n,
plus 1 power n. Then
from the denominator,
we can pull n,
outside means n, to
the power n, outside.
Then you can cancel
the numerator n,
to the power n, then
we have limit n,
goes to infinite 1 over,
we already pulled n, outside
so what left 1 plus 1 over n,
left to the power n, as it is.
Then take n, goes to infinite,
the denominator goes
to e. Then 1 over e,
is less than 1, hence by
ratio test the series is
absolutely convergent.
Great. So then here's
a quicker side,
if we flip these series n,
to the power n, over n,
factorial, then it turned out to
be [NOISE] this flip here
exactly the same idea,
we will get e, which
is greater than 1.
When greater than 1 then
this series is divergent.
Let's do part c.
The given series is summation
and runs from 1 to infinite of n,
squared plus one over 2n,
squared plus 3, outside power
n. Here's our series
involves the nth,
power that means we are
going to take root test.
Solutions here, a_n, equals to n,
squared plus 1 over 2n,
squared plus 3 to the
power n. This means n,
is power, [NOISE] n, is the power
of both numerator as
well as denominator.
Let's take the limit of the nth,
root of a_n, absolute value.
Then next step limit
n, goes to infinite.
Then nth, root to n,
power cancels so we only have n,
squared plus 1 over
2n, squared plus 3.
Then factor the n,
squared out from
numerator as well as
denominator and cancel
the n, squared,
so and then take
the limit we will
have one-half which
is less than 1.
Hence, by root test the series
is absolutely convergent.
Awesome, let's move on.
Example d. Given
series is summation n,
runs from 1 to infinite,
that's a typo,
it should be infinite, of n,
over 3 to the power n. Let's use
root test because we have
the nth, power there.
Then limit n, goes
to infinite and root
of absolute value of a_n,
equals to, limit n,
goes to infinite.
The nth, root and the nth,
power in the denominator cancel,
so we only have nth,
root of numerator n. It's
turned out to be here,
limit n, goes to infinite nth,
root of n, over 3.
I gave you a fact earlier,
so limit n, nth, root of n,
turned out to be 1, and 1
over 3, is less than 1.
Showed you here nth, root of n,
limit here, limit nth,
root of n, turned out to be 1.
Great. We're here, so
then by root test,
the given series is
absolutely convergent.
Awesome. Let's do example,
e. Given series is n,
runs from 2 to infinite
of negative 1 power n,
minus 1 times 1 over
natural log of n,
to the power n. Here,
a_n, equals to negative
1 to the power n,
minus 1 times 1 over ln,
n, to the power n. So [NOISE]
let's apply the root test.
So the limit n, goes
to infinite nth,
root of absolute a_n,
equals to limit n,
tends to infinite nth,
root of simply 1 over ln,
n to the power n,
then the nth, root
and n, n cancel out.
So we only have the limit n,
goes to infinite of 1 over ln,
n. Then take a limit when n,
goes to infinite, ln,
n goes to infinite,
then the reciprocal
of infinite is 0,
which is less than 1 hence,
by root test, the given series
is absolutely convergent.
[BACKGROUND] So great.
So we've finished these
sections. Thank you.
