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PROFESSOR: So we
figured out that you

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can get the book stack,
the overhang of the books,

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to be half the harmonic sum.

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With n books, you
can get out Hn over 2

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where Hn is this harmonic
sum, or harmonic number.

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The question is, how are we
going to estimate or calculate

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what this sum is?

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Now it turns out there is no
simple formula for exactly

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what this sum is, but there
is a simple formula that

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estimates it quite accurately.

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And we get the estimation by
bounding the sum by integrals.

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And so let's look at this
integral method for estimating

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sums.

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Remember what I wanted
was the sum of 1 plus 1/2

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plus 1/3 down to 1 over n.

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So let's form some unit
width rectangles of heights

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equal to the amount that I want.

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So here's a rectangle of height
1, rectangle of height 1/2,

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rectangle of height 1/3.

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I'm going out here, if
you actually count to 8,

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but let's propose that
this is a height 1 over n.

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And what I know is that the
total area of these rectangles

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is actually equal to
the number that I want.

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The total area of
these rectangles

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is the harmonic number.

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And I'm interested in a lower
bound for Hn, because I want

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to know how far I can get out.

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I want a tight lower bound.

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It says Hn is larger
than a certain amount.

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That's the amount that I'm sure
that I can stack out in books.

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So the way I'm going to get a
lower bound on this number Hn

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is by looking at
this curve that goes

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through the corners
of the rectangles.

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And if you check it, that
curve is 1 over x plus 1.

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That is, the point here is--
when x is 0, I'm at 1 over 1.

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When x is 1, I'm at 1/2, the
height of the second rectangle,

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and so on.

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So 1 over x plus
1 is a curve that

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is strictly below the boundaries
of all these rectangles.

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That means that the area under 1
over x plus 1 going from 0 to n

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is a lower bound on Hn,
because it's a lower bound

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on the area of the rectangles.

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So Hn equals the area
of the rectangles.

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It's greater than the
area of 1 over x plus 1,

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which of course is equal
to the integral from 0

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to n of 1 over x plus 1,
which shifting variables is

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the same as the integral from
1 to n plus 1 of 1 over x

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dx, which of course we know is
a natural logarithm of n plus 1.

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So there we have it.

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The overhang that you need
for three books, which

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is Bn greater than
or equal to 3,

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means that Hn has to be
greater than or equal to 6.

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So by this estimate,
I need log of n

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plus 1 greater
than or equal to 6

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in order to get three books
out, that the back end

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of the top book is two books
past the edge of the table

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and the right end of
the furthest out book

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is three book lengths past
the edge of the table.

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So my bound tells me
that I need n books such

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that log of n plus 1 is
greater than or equal to 6.

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Well, exponentiating both
sides, the right-hand side

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becomes e to the 6th.

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And I figure out
that as long as n

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is greater than or equal to e to
the 6th minus 1 books-- rounded

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up, of course, because there's
only-- you can't have fractions

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of a book-- you get an
estimate that with 403 books,

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I can actually get my stack to
stick out three book lengths

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past the edge of the table.

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Well if you do the
actual calculation

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instead of the
estimate, it turns out

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that 227 books are enough.

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So this estimate's a little off.

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But for our
purposes, it tells us

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a dramatic fact, which is that
we know that log of n plus 1

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approaches infinity as
n approaches infinity.

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And that means that,
with enough books,

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I can get out as far as I want.

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You tell me how many book
lengths you want to be out,

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I'll use the log n formula
to calculate how many books I

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need to get that far out.

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So here's an example of
some students in the class

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some years ago decided to
do this as an experiment.

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Now when we used to
do this in class,

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we first tried to do it
with the big, heavy textbook

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that we used.

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And we kept trying to get
them to balance and go out

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over the edge of
the table, and they

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failed, because it turns
out that textbooks are heavy

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and they compress.

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They're not the rigid
rectangular cross sections

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that our model was based on.

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But CD cases work very well.

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They are more rigid.

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They don't compress easily,
and they're very lightweight

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so that they don't
cause problems

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with distortions because
of the size of the stack.

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And so you can actually get CD
cases to stick out pretty far.

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This is an example where
it's 43 CD cases high,

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and the top four
cases are completely

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past the edge of the table.

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The leftmost edge is about
1.81 or 1.91 case lengths

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past the table.

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There's another view of it from
the guy who made the stack.

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And of course, they were
right on the edge of stability

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in trying to get the CDs to
stick out as far as possible.

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So if you notice these
little spaces there,

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in terms of the balancing,
it's really just

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on the brink of falling over.

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If you sneeze at it, it'll tip.

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But if you don't sneeze
at it, it's stable,

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and you get the top
CD out that far.

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So while we're at it, let's
get an upper bound for Hn.

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We just got a lower bound
of Hn, but the same kind

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of logic of using an
integral will give you

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an upper bound for Hn.

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What I do now is I run a
curve from the upper right

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corners of the rectangles.

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And that curve is
simply 1 over x.

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So an upper bound for
the harmonic number

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Hn is the area under 1 over
x out to n plus this 1.

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And so I get an upper bound that
says that the harmonic number

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Hn is less than
the integral from 1

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to n of 1 over x dx plus 1, or
it's equal to 1 plus log of n.

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So combining those
two bounds that I

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got by looking at a curve
that's a lower bound on the area

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and a curve that's an
upper bound on the area

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and integrating, I discover
that Hn is bracketed

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between the natural
log of n plus 1 and 1

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plus the natural log of n.

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Now these two numbers, log of
n plus 1 and 1 plus log of n,

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are very close, and they get
closer and closer as n grows.

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So it turns out that what
we can say pretty accurately

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is that Hn is asymptotically
equal to log n.

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It's approximately
equal to log n.

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And the precise definition
of this tilde symbol

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that I've highlighted in magenta
is called asymptotically equal.

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And the general definition
of asymptotically equal,

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that Hn is asymptotically
equal to log n,

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is that a function f
of n is asymptotically

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equal to a function g of n when
the limit of their quotient

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goes to 1.

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That is to say, as a
multiplicative factor,

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each is within a constant one
of the other in the limit.

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1 plus epsilon of-- 1 plus
or minus epsilon of gn

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is going to bracket
fn, and vice versa.

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Asymptotic equivalents,
or asymptotic equality.

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Let's do an example.

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So the remark would
be, for example,

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that n squared plus
n is asymptotically

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equal to n squared.

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Why?

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Well let's look at the limit
as n approaches infinity of n

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squared plus n over n squared.

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It's the same as
simplifying algebraically

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the limit of 1 over 1 plus n.

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As n approaches infinity,
that term goes to zero.

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Sure enough, the limit is
1, and so these two terms

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are asymptotically equal.

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The idea of asymptotically
equal is that all we care about

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is the high order term.

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The low order terms
will disappear,

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and we're looking at
the principal term that

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controls the growth rate of the
functions when we look at them

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up to asymptotic equivalence.

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So let's step back
for just a moment

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and generalize what
we've done here

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with estimating
the harmonic sum.

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There's a general method called
the integral method where,

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in this particular
case, suppose we

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have a function f, the
positive real valued function

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that's weakly decreasing,
like 1 over x.

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Then let's let S be the sum
from i equals 1 to n of f of i.

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So I was interested where
f of x was 1 over x,

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and I wanted the sum
from 1 over-- 1 to n of 1

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over i, which was the
nth harmonic number.

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And I was comparing that
to the integral from 1 to n

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of f of x, or 1 over
x in our example.

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So if I is the integral and
S is the sum that I want,

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what we can conclude really
is that, in general, the sum

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is bracketed between the
integral plus the first term

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of f of 1 in the sum,
and the integral plus

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the last term of the sum.

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Remember, f is weakly
decreasing, so f of n

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is smaller than f of 1.

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There's a similar theorem
actually, just reverse f of 1

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and f of n, to use
an integral estimate

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to get a bound on a weakly
increasing function.

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And that gives us a
general tool for estimating

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the growth rate of sums.