Real Analysis Lecture 9
Maksim Maydanskiy
Fall 2012
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Limit theorems and their uses
Last time we proved a number of limit theorems that described what happens
to limits when we manipulate sequences termwise. Today we finish discussing
those. They are tools that allow effective limit computations. We then will go
through a collection of basic examples of limits. You can think of those examples
as raw materials for applying the tools, a collection of basic cases to which more
complicated ones should reduce. The result of all this should be able to compute a
significantly greater number of limits than if we could only rely on the definitions.
1.1
Limit theorems.
Recall last time we proved that if sn converges to s and tn converges to t then
an = ksn
converges to a = ks
bn = sn + tn converges to b = s + t
dn = sn tn
converges to d = st.
We have also advertised cn = stnn . But dividing is multiplying by the inverse!
(For us this was, in fact the definition of dividing - fields initially have only addition and multiplication, and division and subtraction are defined on top using
axioms for inverses). In any case, after all that work proving the limit of product
theorem, here is our chance to put it to use.
Proposition 1.1. If sn (with sn 6= 0 for all n) converges to s 6= 0 then s1n converges
to 1s .
Corollary 1.2. If sn converges to s, and tn (with tn =6= 0 for all n) converges to
t 6= 0 then cn = sn /tn converges to c = s/t.
Proof. By Proposition 1.1, t1n converges to 1t . Then by product limit theorem,
sn /tn = sn t1n converges to s 1t = s/t, as wanted.
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Proof of Proposition 1.1. We want 1s − s1n = ssnn−s
< ε, for large N . This is a
s product, so we bound one term and show the other goes to zero. That is we need
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< B for some bound B and |sn − s| < ε/B for all n large enough. The first
sn s
one is equivalent to sn being bounded below by a positive constant. We leave the
proof of this as well as the formal proof of this proposition as an exercise.
Exercise 1.3. Show that is sn 6= 0 converges to s 6= 0, then |sn | is bounded below
by b > 0. Hint: Divide the sequence of |sn | into the initial part which is finite and
hence has a minimum, and the infinite “tail” which is sufficiently close to |s| to be
bounded away from 0. Combine the two bounds - for the starting part and for the
tail- to finish.(Or look at Eaxmple 6 p.41 of Ross).
Exercise 1.4. Write the formal proof of Proposition 1.1.
1.2
Examples.
Ok, that’s quite a bunch of theorems. As payoff, let’s show that the example we
did last time becomes easy.
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Example 1.5. Prove that sn = 2nn2 +n converges to 12 using the limit theorems.
Hint: Divide the nominator and the denominator by n2 .
Many similar statements become routine. Note that we used limn→∞ n1 = 0 as
raw material and the tools of limit theorems gave us the rest. This is sufficient to
tackle all limits where sn is a ratio of two polynomials. However to handle more
complicated limits we need more raw material. Before we get to that, let’s introduce a bit more language, which I feel would make the discussion less awkward.
Definition 1.6. We say that a sequence sn diverges to positive infinity if
for every M there exists an N such that for all n > N we have
sn > M .
We write limn→∞ = +∞.
This definition is very similar to the definition of convergence; in fact we can
think of it as expressing the notion that sn converges to a “point at infinity”. There
is a way to make this precise using the notion of topological space, but we shall
not pursue this here. We can think of it as a useful heuristic.
Ross uses “for every M > 0” in his definition, presumably to preserve the
familiar rhythm of “for each ...¿0 exists ”; convince yourself that the two versions
are equivalent. Also aonvince yourself that “for all M > 10000” would work just
as well.
We have also
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Definition 1.7. We say that a sequence sn diverges to negative infinity if
for every m there exists an N such that for all n > N we have sn <
m.
Exercise 1.8. Show that sn converges to positive infinity if and only if −sn diverges to negative infinity.
We say that sn has a limit if it converges, or diverges to positive or negative
infinity.
The reason this language is useful is the following proposition.
Proposition 1.9. Suppose sn is a sequence of positive numbers. Then sn diverges
to positive infinity if and only if s1n converges to 0.
Proof. Note that sn > x is the same as s1n < x1 .
Suppose sn diverges to plus infinity. So given ε > 0 we can take M = 1ε and
there is N such that sn > M for all n > N . Then s1n < ε, and so s1n converges to
0.
Conversely, suppose s1n converges to 0. So given M > 0 we can take ε = M1
and there is N such that s1n < ε for all n > N . Then sn > M , and so sn diverges
to plus infinity.
This allows us to translate statements like limn→∞
n = +∞, which I find more immediately obvious.
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n2 +n
= 0 to limn→∞ n2 +
Example 1.10. lim np = +∞ for p > 0.
For integer p we can prove this by induction, applying the “limit of products”
theorem (the divergent version). We have not yet defined what np means for noninteger p,∗ but we shall assume that such a number xp is meaningful for all x > 0
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and p > 0. We shall also assume things like (xp ) p = (x p )p = x and x > y implies
xp > y p for all p > 0. Then the proof of the limit becomes almost obvious, as
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np > M is the same as n > M p .
Next time we shall finish proving limit theorems for divergent sequences and
add a couple more examples to our arsenal.
∗
To be completely formal, we have not defined np for integer p either, but I hope it is clear that
we could do that by induction (np+1 = np · n).
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