
English: 
Hi.
Welcome back to recitation.
In class, you've been learning
about convergence tests for
infinite series.
I have here three examples of
series that I happen to like.
So the first one is the sum from
n equals 0 to infinity of
5n plus 2 divided by
n cubed plus 1.
The second one is the sum from
n equals 1 to infinity of the
quantity 1 plus the square root
of 5 over 2, all that to
the nth power.
And the third one is the sum
from n equals 1 to infinity of
the natural log of n divided
by n squared.
So what I'd like you to do for
each of these three series is
to figure out whether or
not they converge.
So whether they converge,
or whether they diverge.
So why don't you pause the
video, take some time to work
that out, come back, and we
can work on it together.

English: 
Hi.
Welcome back to recitation.
In class, you've been learning
about convergence tests for
infinite series.
I have here three examples of
series that I happen to like.
So the first one is the sum from
n equals 0 to infinity of
5n plus 2 divided by
n cubed plus 1.
The second one is the sum from
n equals 1 to infinity of the
quantity 1 plus the square root
of 5 over 2, all that to
the nth power.
And the third one is the sum
from n equals 1 to infinity of
the natural log of n divided
by n squared.
So what I'd like you to do for
each of these three series is
to figure out whether or
not they converge.
So whether they converge,
or whether they diverge.
So why don't you pause the
video, take some time to work
that out, come back, and we
can work on it together.

English: 
Welcome back.
Hopefully you had some luck
working on these series.
So let's talk about
how we do them.
So I'll start with the
first one here.
So this first one, the function
that we're, that
we're summing is this 5n plus
2 divided by n cubed plus 3.
And so you want to know, does
that sum of that expression,
of n goes from 1 to infinity,
does that converge?
Does it reach some
finite value?
Or does it diverge?
Does it either, you know,
oscillate, or go to infinity,
or something like that?
So for one thing you can always
do, is something like a
comparison test. And so what you
want to do there, is you
want to extract the information
from the
expression for the summand.
You want to figure out, you
know, about how big this is.
What are the important
parts of it?
So in this case, you
have a ratio.
And so when you have a ratio,
what you want to do, is you
want to look at what's the
magnitude of the top, and
what's the magnitude
of the bottom.
So roughly speaking, the
magnitude of the top is of the

English: 
Welcome back.
Hopefully you had some luck
working on these series.
So let's talk about
how we do them.
So I'll start with the
first one here.
So this first one, the function
that we're, that
we're summing is this 5n plus
2 divided by n cubed plus 3.
And so you want to know, does
that sum of that expression,
of n goes from 1 to infinity,
does that converge?
Does it reach some
finite value?
Or does it diverge?
Does it either, you know,
oscillate, or go to infinity,
or something like that?
So for one thing you can always
do, is something like a
comparison test. And so what you
want to do there, is you
want to extract the information
from the
expression for the summand.
You want to figure out, you
know, about how big this is.
What are the important
parts of it?
So in this case, you
have a ratio.
And so when you have a ratio,
what you want to do, is you
want to look at what's the
magnitude of the top, and
what's the magnitude
of the bottom.
So roughly speaking, the
magnitude of the top is of the

English: 
order of n.
Right?
It's 5n.
The 5 is just a constant
multiple.
That's not going to make
a whole big difference.
And the plus 2 is just much,
much, much smaller than the n
when n gets very, very large.
So the, we can say that the
order of magnitude of the top
here is of the order of n, and
similarly, the order of
magnitude of the bottom is
of the order of n cubed.
Right?
The plus 1 is much, much, much
smaller than the n cubed, so
it's not, it's almost
insignificant.
So roughly speaking, we should
expect this to be of the same
behavior as if we just
had n over n cubed,
which is 1 over n squared.
And we know that 1 over n
squared, the series 1 over n
squared from n equals 1 to
infinity, converges.
So that's sort of just
a sort of a way of
thinking about this.
And we can formalize it using
the limit comparison test.

English: 
order of n.
Right?
It's 5n.
The 5 is just a constant
multiple.
That's not going to make
a whole big difference.
And the plus 2 is just much,
much, much smaller than the n
when n gets very, very large.
So the, we can say that the
order of magnitude of the top
here is of the order of n, and
similarly, the order of
magnitude of the bottom is
of the order of n cubed.
Right?
The plus 1 is much, much, much
smaller than the n cubed, so
it's not, it's almost
insignificant.
So roughly speaking, we should
expect this to be of the same
behavior as if we just
had n over n cubed,
which is 1 over n squared.
And we know that 1 over n
squared, the series 1 over n
squared from n equals 1 to
infinity, converges.
So that's sort of just
a sort of a way of
thinking about this.
And we can formalize it using
the limit comparison test.

English: 
So the idea here for part A is
to use the limit comparison
test. So what we want
to compute--
so I'm going to put a 5 in.
So I'm going to make it 5 over
n squared, because we
had this 5 up here.
So we have an is equal to 5n
plus 2 over n cubed plus 1.
And I want to limit
compare it with--
I'm going to compare my series
with the series sum from n
equals 1 to infinity of
5 over n squared.
Now--
I'm talking about
this 5, right?
This 5 doesn't matter.
If this series-- if 1 over n
squared, that sum converges
then 5 over n squared, that
sum converges to exactly 5
times as much.
And if, you know, if I wrote a
divergent thing here, and then
multiplied it by 5, the
result multiplied by
5 would still diverge.

English: 
So the idea here for part A is
to use the limit comparison
test. So what we want
to compute--
so I'm going to put a 5 in.
So I'm going to make it 5 over
n squared, because we
had this 5 up here.
So we have an is equal to 5n
plus 2 over n cubed plus 1.
And I want to limit
compare it with--
I'm going to compare my series
with the series sum from n
equals 1 to infinity of
5 over n squared.
Now--
I'm talking about
this 5, right?
This 5 doesn't matter.
If this series-- if 1 over n
squared, that sum converges
then 5 over n squared, that
sum converges to exactly 5
times as much.
And if, you know, if I wrote a
divergent thing here, and then
multiplied it by 5, the
result multiplied by
5 would still diverge.

English: 
So this constant multiple
isn't going to matter.
And I'm just choosing, I'm just
putting the 5 in there
because of this 5 we
saw over here.
So OK.
So let's select work
it out, then.
So the limit comparison
test says--
so we look at the limit as n
goes to infinity of the ratio
of the two things that
we're interested in.
So in this case, this is 5n and
plus 2 over n cubed plus
1, divided by 5 over
n squared.
And you can--
OK.
So this is a ratio of two
ratios, so we can rewrite it
by multiplying upstairs, and we
get that this is equal to
the limit as n goes to infinity
of 5n cubed plus 2 n
squared over 5n cubed plus 5.
And so we've seen before, when
we were dealing with limits,

English: 
So this constant multiple
isn't going to matter.
And I'm just choosing, I'm just
putting the 5 in there
because of this 5 we
saw over here.
So OK.
So let's select work
it out, then.
So the limit comparison
test says--
so we look at the limit as n
goes to infinity of the ratio
of the two things that
we're interested in.
So in this case, this is 5n and
plus 2 over n cubed plus
1, divided by 5 over
n squared.
And you can--
OK.
So this is a ratio of two
ratios, so we can rewrite it
by multiplying upstairs, and we
get that this is equal to
the limit as n goes to infinity
of 5n cubed plus 2 n
squared over 5n cubed plus 5.
And so we've seen before, when
we were dealing with limits,

English: 
that when you have a ratio of
two polynomials, as the
variable goes to infinity, the
limit is what you get just by
comparing the leading terms. So
in this case, we just have
to look at 5n cubed over 5n
cubed, and that's indeed 1.
OK, so this is 1.
So by the limit comparison test,
our series, the sum of
this an, converges if and only
if this series converges.
And we already said
that we know sum 1
over n squared converges.
So by the limit comparison
test, since sum 1 over n
squared converges, our
series converges.
Great.
OK.
So that's the first one.
Let's look at the second one.
So the second one here is the
sum from n equals 1 to
infinity of 1 plus the square
root of 5 over 2.
That whole thing to
the nth power.
Now, if you look at this, the
thing you should recognize
this as, is just a particular
geometric series.
Right?

English: 
that when you have a ratio of
two polynomials, as the
variable goes to infinity, the
limit is what you get just by
comparing the leading terms. So
in this case, we just have
to look at 5n cubed over 5n
cubed, and that's indeed 1.
OK, so this is 1.
So by the limit comparison test,
our series, the sum of
this an, converges if and only
if this series converges.
And we already said
that we know sum 1
over n squared converges.
So by the limit comparison
test, since sum 1 over n
squared converges, our
series converges.
Great.
OK.
So that's the first one.
Let's look at the second one.
So the second one here is the
sum from n equals 1 to
infinity of 1 plus the square
root of 5 over 2.
That whole thing to
the nth power.
Now, if you look at this, the
thing you should recognize
this as, is just a particular
geometric series.
Right?

English: 
This is, if you were to replace
1 plus the square root
of 5 over 2 with x, this is just
x plus x squared plus x
cubed plus dot dot
dot dot dot.
It's just a geometric series
with constant ratio--
well, x .
1 plus the square root
of 5 over 2.
So we know exactly when
geometric series converge.
They converge exactly when the
constant ratio is between
minus 1 and 1.
So bigger than minus
1 and less than 1.
So this series converges, then,
if and only if 1 plus
the square root of 5 over 2
is between minus 1 and 1.
So then we just have to think
about, is this number between
minus 1 and 1.
And OK.
So this is not that
hard to do.
Square root of 5 is bigger than
2, so 1 plus the square
root of 5 is bigger than 3, so
1 plus the square root of 5
over 2 is bigger than 3/2.
So 3/2 is bigger than 1.
So this common ratio in this
series is bigger than 1.
So the terms of this series
are blowing ip.

English: 
This is, if you were to replace
1 plus the square root
of 5 over 2 with x, this is just
x plus x squared plus x
cubed plus dot dot
dot dot dot.
It's just a geometric series
with constant ratio--
well, x .
1 plus the square root
of 5 over 2.
So we know exactly when
geometric series converge.
They converge exactly when the
constant ratio is between
minus 1 and 1.
So bigger than minus
1 and less than 1.
So this series converges, then,
if and only if 1 plus
the square root of 5 over 2
is between minus 1 and 1.
So then we just have to think
about, is this number between
minus 1 and 1.
And OK.
So this is not that
hard to do.
Square root of 5 is bigger than
2, so 1 plus the square
root of 5 is bigger than 3, so
1 plus the square root of 5
over 2 is bigger than 3/2.
So 3/2 is bigger than 1.
So this common ratio in this
series is bigger than 1.
So the terms of this series
are blowing ip.

English: 
You know, when you, when n gets
bigger and bigger, you're
adding larger and larger
numbers here.
This is blowing up.
It's a divergent geometric
series.
So this series does
not converge.
So b, I'll just write
that here.
B diverges, because it's
geometric with common ratio
bigger than 1.
So that's the reason that
part B diverges.
Finally, we're left
with question C.
So I'm going to come over and
write it over here again, so
we can see it.
So part C asks the sum for n
equals 1 to infinity of log n
over n squared.
So this one's a little trickier,
and it requires a
little bit more thought.
The thing to-- do let's
start just by--

English: 
You know, when you, when n gets
bigger and bigger, you're
adding larger and larger
numbers here.
This is blowing up.
It's a divergent geometric
series.
So this series does
not converge.
So b, I'll just write
that here.
B diverges, because it's
geometric with common ratio
bigger than 1.
So that's the reason that
part B diverges.
Finally, we're left
with question C.
So I'm going to come over and
write it over here again, so
we can see it.
So part C asks the sum for n
equals 1 to infinity of log n
over n squared.
So this one's a little trickier,
and it requires a
little bit more thought.
The thing to-- do let's
start just by--

English: 
it can't be solved just by a
straightforward application of
the limit comparison test
that we've learned.
So we need to think a little bit
more about what ways could
we solve it.
So one thing to remember here
is, is we should think about,
what are the magnitudes
of these things?
So we know that sum 1 over
n squared converges.
But log n is a function
that grows.
So this individual term is
bigger than 1 over n squared.
So we can't just compare
it to 1 over n squared.
But log n grows very,
very slowly.
How slowly?
Well, it grows more slowly
than any power of x.
Right?
So, or-- sorry-- any power of
n in this case, because the
variable is n.
So if you remember, we know,
we've shown, that the limit of

English: 
it can't be solved just by a
straightforward application of
the limit comparison test
that we've learned.
So we need to think a little bit
more about what ways could
we solve it.
So one thing to remember here
is, is we should think about,
what are the magnitudes
of these things?
So we know that sum 1 over
n squared converges.
But log n is a function
that grows.
So this individual term is
bigger than 1 over n squared.
So we can't just compare
it to 1 over n squared.
But log n grows very,
very slowly.
How slowly?
Well, it grows more slowly
than any power of x.
Right?
So, or-- sorry-- any power of
n in this case, because the
variable is n.
So if you remember, we know,
we've shown, that the limit of

English: 
ln x over x to the p as x goes
to infinity is equal to 0 for
any positive p.
So ln, log n, ln n, is going to
infinity, but it's going to
infinity much slower than
any power of n.
In this case.
Or x, down here.
So OK.
So what does that mean?
Well, one thing you could do,
is you could say, oh, OK.
So ln n over n, that's
getting really small.
And then what we're left
with is 1 over n.
So you can say, OK.
So this is much smaller
than 1 over n.
The problem is that the
sum 1 over n diverges.
Yeah?
So that doesn't help
us, really, right?
So we've shown this is bigger
than the sum 1 over n squared,
which converges, and it's
smaller than the sum 1 over n,
which diverges.
But that still doesn't
tell us, you know?

English: 
ln x over x to the p as x goes
to infinity is equal to 0 for
any positive p.
So ln, log n, ln n, is going to
infinity, but it's going to
infinity much slower than
any power of n.
In this case.
Or x, down here.
So OK.
So what does that mean?
Well, one thing you could do,
is you could say, oh, OK.
So ln n over n, that's
getting really small.
And then what we're left
with is 1 over n.
So you can say, OK.
So this is much smaller
than 1 over n.
The problem is that the
sum 1 over n diverges.
Yeah?
So that doesn't help
us, really, right?
So we've shown this is bigger
than the sum 1 over n squared,
which converges, and it's
smaller than the sum 1 over n,
which diverges.
But that still doesn't
tell us, you know?

English: 
It could be something that,
you know, there's a lot of
room bigger than a particular
convergent series, and smaller
than a particular divergent
series.
And in particular, there are
both convergent and divergent
series in between.
So we still need, we need
either something that
converges that our thing is
less than, or we need
something that diverges that
our thing is bigger than.
Right?
If we can bound our series above
by something convergent,
then our series converges.
Because [UNINTELLIGIBLE]
has positive terms.
This is important.
Or if we can bound it below by
something that diverges, then
we would know it diverges.
And so far, we haven't
been able to do that.
But maybe we can think of a-- so
I said we could write this,
a second ago, I said we could
write ln n over n squared is
equal to ln n and over
n times 1 over n.
So that was what we just said
a minute ago, at which we

English: 
It could be something that,
you know, there's a lot of
room bigger than a particular
convergent series, and smaller
than a particular divergent
series.
And in particular, there are
both convergent and divergent
series in between.
So we still need, we need
either something that
converges that our thing is
less than, or we need
something that diverges that
our thing is bigger than.
Right?
If we can bound our series above
by something convergent,
then our series converges.
Because [UNINTELLIGIBLE]
has positive terms.
This is important.
Or if we can bound it below by
something that diverges, then
we would know it diverges.
And so far, we haven't
been able to do that.
But maybe we can think of a-- so
I said we could write this,
a second ago, I said we could
write ln n over n squared is
equal to ln n and over
n times 1 over n.
So that was what we just said
a minute ago, at which we

English: 
showed is eventually
less than 1 over n.
So this is true, but it wasn't
useful, because the sum 1 over
n diverges.
But maybe we can, we can
do something even
a little more tricky.
Because here we saw that x, we
could use it in the base, we
could use any power of the
variable, any positive power.
So here, we tried it with
the power n to the 1.
We tried to split this 2 as 1
plus 1, and we kept one of the
n's to knock out the log
n, and we kept the
other n over here.
But we don't need a whole
power of n to
knock out the log n.
Any positive power
of n would do.
So in particular, we could split
this using, say, just a
small power of n, even smaller
than the first power here, and
leave more of it over here.
So we also know, we also have
that ln n over n squared is
equal to, for example,
ln n over--

English: 
showed is eventually
less than 1 over n.
So this is true, but it wasn't
useful, because the sum 1 over
n diverges.
But maybe we can, we can
do something even
a little more tricky.
Because here we saw that x, we
could use it in the base, we
could use any power of the
variable, any positive power.
So here, we tried it with
the power n to the 1.
We tried to split this 2 as 1
plus 1, and we kept one of the
n's to knock out the log
n, and we kept the
other n over here.
But we don't need a whole
power of n to
knock out the log n.
Any positive power
of n would do.
So in particular, we could split
this using, say, just a
small power of n, even smaller
than the first power here, and
leave more of it over here.
So we also know, we also have
that ln n over n squared is
equal to, for example,
ln n over--

English: 
well, you know, for example,
n to the 1/2 times 1
over n to the 3/2.
And now we know that ln n over
n to the 1/2 goes to 0 as n
gets large.
So this is thing is
getting smaller
and smaller and smaller.
So as this gets smaller, we have
that this has to be less
than 1 over n to the 3/2.
So this thing that we're adding
up here is smaller than
1 over n to the 3/2.
Well, what's the significance
of that?
Well, we know 1 over n to the
3/2, that series converges.
Yeah?
We know the sum 1 over n to
the 3/2 from n equals 1 to
infinity converges, because
3/2 is bigger than 1.
OK?
So what does that mean?
Well, we have our series, and
we've shown that the terms of

English: 
well, you know, for example,
n to the 1/2 times 1
over n to the 3/2.
And now we know that ln n over
n to the 1/2 goes to 0 as n
gets large.
So this is thing is
getting smaller
and smaller and smaller.
So as this gets smaller, we have
that this has to be less
than 1 over n to the 3/2.
So this thing that we're adding
up here is smaller than
1 over n to the 3/2.
Well, what's the significance
of that?
Well, we know 1 over n to the
3/2, that series converges.
Yeah?
We know the sum 1 over n to
the 3/2 from n equals 1 to
infinity converges, because
3/2 is bigger than 1.
OK?
So what does that mean?
Well, we have our series, and
we've shown that the terms of

English: 
our series are eventually
bounded below 1
over n to the 3/2.
And we know that the sum of 1
over n to the 3/2 converges,
so our series is bounded above
by a convergent series.
So whenever you have a series
bounded above, a series of
non-negative terms bounded
above, by a convergent series,
that means your series
also has to converge.
OK, so this converges.
So it converges by a comparison
to this other series.
Using this cute trick that we
can replace a log with any
small positive power of n that
happens to be convenient.
And of course, if you wanted to,
maybe you could have made
this 1/2 a 1/10 and that would
have been fine, or you
could've even made it a 9/10,
and then you would here be
left with n to the 11/10, and
that would still be OK.
Because that would be 11/10, and
11/10, then it would still
be bigger than 1.
All right.
So this is a nice use of a, of
a comparison test. We didn't

English: 
our series are eventually
bounded below 1
over n to the 3/2.
And we know that the sum of 1
over n to the 3/2 converges,
so our series is bounded above
by a convergent series.
So whenever you have a series
bounded above, a series of
non-negative terms bounded
above, by a convergent series,
that means your series
also has to converge.
OK, so this converges.
So it converges by a comparison
to this other series.
Using this cute trick that we
can replace a log with any
small positive power of n that
happens to be convenient.
And of course, if you wanted to,
maybe you could have made
this 1/2 a 1/10 and that would
have been fine, or you
could've even made it a 9/10,
and then you would here be
left with n to the 11/10, and
that would still be OK.
Because that would be 11/10, and
11/10, then it would still
be bigger than 1.
All right.
So this is a nice use of a, of
a comparison test. We didn't

English: 
use exactly the limit comparison
test as it was
described in lecture, but it's a
very closely related process
that we went through to show
that this second series--
sorry, this third series--
also converges.
So I'll end there.

English: 
use exactly the limit comparison
test as it was
described in lecture, but it's a
very closely related process
that we went through to show
that this second series--
sorry, this third series--
also converges.
So I'll end there.
