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Hi.

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Welcome back to recitation.

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In class, you've been learning
about convergence tests for

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infinite series.

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I have here three examples of
series that I happen to like.

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So the first one is the sum from
n equals 0 to infinity of

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5n plus 2 divided by
n cubed plus 1.

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The second one is the sum from
n equals 1 to infinity of the

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quantity 1 plus the square root
of 5 over 2, all that to

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the nth power.

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And the third one is the sum
from n equals 1 to infinity of

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the natural log of n divided
by n squared.

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So what I'd like you to do for
each of these three series is

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to figure out whether or
not they converge.

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So whether they converge,
or whether they diverge.

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So why don't you pause the
video, take some time to work

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that out, come back, and we
can work on it together.

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Welcome back.

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Hopefully you had some luck
working on these series.

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So let's talk about
how we do them.

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So I'll start with the
first one here.

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So this first one, the function
that we're, that

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we're summing is this 5n plus
2 divided by n cubed plus 3.

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And so you want to know, does
that sum of that expression,

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of n goes from 1 to infinity,
does that converge?

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Does it reach some
finite value?

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Or does it diverge?

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Does it either, you know,
oscillate, or go to infinity,

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or something like that?

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So for one thing you can always
do, is something like a

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comparison test. And so what you
want to do there, is you

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want to extract the information
from the

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expression for the summand.

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You want to figure out, you
know, about how big this is.

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What are the important
parts of it?

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So in this case, you
have a ratio.

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And so when you have a ratio,
what you want to do, is you

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want to look at what's the
magnitude of the top, and

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what's the magnitude
of the bottom.

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So roughly speaking, the
magnitude of the top is of the

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order of n.

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Right?

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It's 5n.

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The 5 is just a constant
multiple.

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That's not going to make
a whole big difference.

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And the plus 2 is just much,
much, much smaller than the n

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when n gets very, very large.

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So the, we can say that the
order of magnitude of the top

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here is of the order of n, and
similarly, the order of

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magnitude of the bottom is
of the order of n cubed.

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Right?

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The plus 1 is much, much, much
smaller than the n cubed, so

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it's not, it's almost
insignificant.

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So roughly speaking, we should
expect this to be of the same

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behavior as if we just
had n over n cubed,

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which is 1 over n squared.

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And we know that 1 over n
squared, the series 1 over n

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squared from n equals 1 to
infinity, converges.

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So that's sort of just
a sort of a way of

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thinking about this.

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And we can formalize it using
the limit comparison test.

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So the idea here for part A is
to use the limit comparison

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test. So what we want
to compute--

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so I'm going to put a 5 in.

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So I'm going to make it 5 over
n squared, because we

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had this 5 up here.

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So we have an is equal to 5n
plus 2 over n cubed plus 1.

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And I want to limit
compare it with--

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I'm going to compare my series
with the series sum from n

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equals 1 to infinity of
5 over n squared.

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Now--

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I'm talking about
this 5, right?

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This 5 doesn't matter.

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If this series-- if 1 over n
squared, that sum converges

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then 5 over n squared, that
sum converges to exactly 5

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times as much.

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And if, you know, if I wrote a
divergent thing here, and then

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multiplied it by 5, the
result multiplied by

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5 would still diverge.

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So this constant multiple
isn't going to matter.

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And I'm just choosing, I'm just
putting the 5 in there

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because of this 5 we
saw over here.

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So OK.

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So let's select work
it out, then.

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So the limit comparison
test says--

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so we look at the limit as n
goes to infinity of the ratio

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of the two things that
we're interested in.

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So in this case, this is 5n and
plus 2 over n cubed plus

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1, divided by 5 over
n squared.

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And you can--

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OK.

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So this is a ratio of two
ratios, so we can rewrite it

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by multiplying upstairs, and we
get that this is equal to

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the limit as n goes to infinity
of 5n cubed plus 2 n

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squared over 5n cubed plus 5.

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And so we've seen before, when
we were dealing with limits,

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that when you have a ratio of
two polynomials, as the

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variable goes to infinity, the
limit is what you get just by

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comparing the leading terms. So
in this case, we just have

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to look at 5n cubed over 5n
cubed, and that's indeed 1.

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OK, so this is 1.

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So by the limit comparison test,
our series, the sum of

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this an, converges if and only
if this series converges.

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And we already said
that we know sum 1

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over n squared converges.

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So by the limit comparison
test, since sum 1 over n

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squared converges, our
series converges.

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Great.

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OK.

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So that's the first one.

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Let's look at the second one.

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So the second one here is the
sum from n equals 1 to

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infinity of 1 plus the square
root of 5 over 2.

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That whole thing to
the nth power.

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Now, if you look at this, the
thing you should recognize

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this as, is just a particular
geometric series.

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Right?

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This is, if you were to replace
1 plus the square root

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of 5 over 2 with x, this is just
x plus x squared plus x

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cubed plus dot dot
dot dot dot.

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It's just a geometric series
with constant ratio--

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well, x .

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1 plus the square root
of 5 over 2.

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So we know exactly when
geometric series converge.

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They converge exactly when the
constant ratio is between

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minus 1 and 1.

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So bigger than minus
1 and less than 1.

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So this series converges, then,
if and only if 1 plus

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the square root of 5 over 2
is between minus 1 and 1.

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So then we just have to think
about, is this number between

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minus 1 and 1.

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And OK.

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So this is not that
hard to do.

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Square root of 5 is bigger than
2, so 1 plus the square

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root of 5 is bigger than 3, so
1 plus the square root of 5

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over 2 is bigger than 3/2.

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So 3/2 is bigger than 1.

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So this common ratio in this
series is bigger than 1.

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So the terms of this series
are blowing ip.

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You know, when you, when n gets
bigger and bigger, you're

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adding larger and larger
numbers here.

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This is blowing up.

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It's a divergent geometric
series.

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So this series does
not converge.

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So b, I'll just write
that here.

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B diverges, because it's
geometric with common ratio

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bigger than 1.

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So that's the reason that
part B diverges.

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Finally, we're left
with question C.

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So I'm going to come over and
write it over here again, so

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we can see it.

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So part C asks the sum for n
equals 1 to infinity of log n

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over n squared.

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So this one's a little trickier,
and it requires a

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little bit more thought.

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The thing to-- do let's
start just by--

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it can't be solved just by a
straightforward application of

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the limit comparison test
that we've learned.

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So we need to think a little bit
more about what ways could

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we solve it.

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So one thing to remember here
is, is we should think about,

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what are the magnitudes
of these things?

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So we know that sum 1 over
n squared converges.

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But log n is a function
that grows.

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So this individual term is
bigger than 1 over n squared.

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So we can't just compare
it to 1 over n squared.

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But log n grows very,
very slowly.

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How slowly?

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Well, it grows more slowly
than any power of x.

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Right?

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So, or-- sorry-- any power of
n in this case, because the

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variable is n.

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So if you remember, we know,
we've shown, that the limit of

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ln x over x to the p as x goes
to infinity is equal to 0 for

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any positive p.

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So ln, log n, ln n, is going to
infinity, but it's going to

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infinity much slower than
any power of n.

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In this case.

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Or x, down here.

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So OK.

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So what does that mean?

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Well, one thing you could do,
is you could say, oh, OK.

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So ln n over n, that's
getting really small.

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And then what we're left
with is 1 over n.

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So you can say, OK.

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So this is much smaller
than 1 over n.

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The problem is that the
sum 1 over n diverges.

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Yeah?

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So that doesn't help
us, really, right?

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So we've shown this is bigger
than the sum 1 over n squared,

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which converges, and it's
smaller than the sum 1 over n,

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which diverges.

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But that still doesn't
tell us, you know?

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It could be something that,
you know, there's a lot of

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room bigger than a particular
convergent series, and smaller

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than a particular divergent
series.

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And in particular, there are
both convergent and divergent

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series in between.

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So we still need, we need
either something that

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converges that our thing is
less than, or we need

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something that diverges that
our thing is bigger than.

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Right?

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If we can bound our series above
by something convergent,

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then our series converges.

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Because [UNINTELLIGIBLE]

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has positive terms.
This is important.

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Or if we can bound it below by
something that diverges, then

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we would know it diverges.

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And so far, we haven't
been able to do that.

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But maybe we can think of a-- so
I said we could write this,

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a second ago, I said we could
write ln n over n squared is

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equal to ln n and over
n times 1 over n.

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So that was what we just said
a minute ago, at which we

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showed is eventually
less than 1 over n.

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So this is true, but it wasn't
useful, because the sum 1 over

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n diverges.

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But maybe we can, we can
do something even

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a little more tricky.

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Because here we saw that x, we
could use it in the base, we

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could use any power of the
variable, any positive power.

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So here, we tried it with
the power n to the 1.

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We tried to split this 2 as 1
plus 1, and we kept one of the

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n's to knock out the log
n, and we kept the

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other n over here.

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But we don't need a whole
power of n to

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knock out the log n.

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Any positive power
of n would do.

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So in particular, we could split
this using, say, just a

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small power of n, even smaller
than the first power here, and

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leave more of it over here.

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So we also know, we also have
that ln n over n squared is

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equal to, for example,
ln n over--

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well, you know, for example,
n to the 1/2 times 1

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over n to the 3/2.

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And now we know that ln n over
n to the 1/2 goes to 0 as n

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gets large.

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So this is thing is
getting smaller

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and smaller and smaller.

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So as this gets smaller, we have
that this has to be less

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than 1 over n to the 3/2.

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So this thing that we're adding
up here is smaller than

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1 over n to the 3/2.

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00:12:33,185 --> 00:12:34,550
Well, what's the significance
of that?

244
00:12:34,585 --> 00:12:38,600
Well, we know 1 over n to the
3/2, that series converges.

245
00:12:38,635 --> 00:12:39,330
Yeah?

246
00:12:39,365 --> 00:12:46,990
We know the sum 1 over n to
the 3/2 from n equals 1 to

247
00:12:47,025 --> 00:12:54,050
infinity converges, because
3/2 is bigger than 1.

248
00:12:54,085 --> 00:12:54,620
OK?

249
00:12:54,655 --> 00:12:55,370
So what does that mean?

250
00:12:55,405 --> 00:12:59,130
Well, we have our series, and
we've shown that the terms of

251
00:12:59,165 --> 00:13:02,660
our series are eventually
bounded below 1

252
00:13:02,695 --> 00:13:04,290
over n to the 3/2.

253
00:13:04,325 --> 00:13:08,150
And we know that the sum of 1
over n to the 3/2 converges,

254
00:13:08,185 --> 00:13:12,000
so our series is bounded above
by a convergent series.

255
00:13:12,035 --> 00:13:14,510
So whenever you have a series
bounded above, a series of

256
00:13:14,545 --> 00:13:17,850
non-negative terms bounded
above, by a convergent series,

257
00:13:17,885 --> 00:13:21,460
that means your series
also has to converge.

258
00:13:21,495 --> 00:13:24,388
OK, so this converges.

259
00:13:24,423 --> 00:13:27,230
So it converges by a comparison

260
00:13:27,265 --> 00:13:28,340
to this other series.

261
00:13:28,375 --> 00:13:32,980
Using this cute trick that we
can replace a log with any

262
00:13:33,015 --> 00:13:36,840
small positive power of n that
happens to be convenient.

263
00:13:36,875 --> 00:13:39,290
And of course, if you wanted to,
maybe you could have made

264
00:13:39,325 --> 00:13:42,540
this 1/2 a 1/10 and that would
have been fine, or you

265
00:13:42,575 --> 00:13:45,720
could've even made it a 9/10,
and then you would here be

266
00:13:45,755 --> 00:13:49,280
left with n to the 11/10, and
that would still be OK.

267
00:13:49,315 --> 00:13:53,030
Because that would be 11/10, and
11/10, then it would still

268
00:13:53,065 --> 00:13:54,540
be bigger than 1.

269
00:13:54,575 --> 00:13:55,900
All right.

270
00:13:55,935 --> 00:13:59,810
So this is a nice use of a, of
a comparison test. We didn't

271
00:13:59,845 --> 00:14:02,300
use exactly the limit comparison
test as it was

272
00:14:02,335 --> 00:14:06,220
described in lecture, but it's a
very closely related process

273
00:14:06,255 --> 00:14:10,020
that we went through to show
that this second series--

274
00:14:10,055 --> 00:14:11,540
sorry, this third series--

275
00:14:11,575 --> 00:14:13,480
also converges.

276
00:14:13,515 --> 00:14:15,330
So I'll end there.