Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 9 July 2014

Lecture 10

Sequences

Let Y be a set. A sequence (y1,y2,y3,) in Y is a function >0 Y n yn.

Let Y. Let (y1,y2,) be a sequence in Y. The sequence (y1,y2,) is increasing if (y1,y2,) satisfies if i>0 then yiyi+1. The sequence (y1,y2,) is decreasing if (y1,y2,) satisfies if i>0 then yiyi+1. A sequence (an) is monotone if (an) is increasing or decreasing.

Let Y be a metric space. Let (y1,y2,) be a sequence in Y. The sequence is bounded if the set {y1,y2,} is bounded.

The sequence (y1,y2,) is contractive if (y1,y2,) satisfies: There exists α(0,1) such that if i>0 then d(yi,yi+1)αd(yi-1,yi).

The sequence (y1,y2,) is Cauchy if (y1,y2,) satisfies if ε>0 then there exists N>0 such that if m,n>0 and m>N and n>N then d(ym,yn)<ε.

Let Y. The sequence (y1,y2,) converges to if (y1,y2,) satisfies if ε>0 then there exists N>0 such that if n>0 and n>N then d(yn,)ε.

Let (y1,y2,) be a sequence in (or more generally, any totally ordered set with the order topology). The upper limit of (y1,y2,) is limsupyn= limn sup{yn,yn+1,}. The lower limit of (y1,y2,) is liminfyn= limn inf{yn,yn+1,}.

If yn=(-1)n(1-1n) then limsupyn=1 andliminfyn=-1.

Let (y1,y2,) be a sequence in . Then

(a) limsupyn=sup{cluster points of(y1,y2,)}, and
(b) liminfyn=inf{cluster points of(y1,y2,)}.

Boundedness, sup, inf, limsup and liminf

Let an=lognn.

(a) Show that an is bounded.
(b) Find N>0 such that an is decreasing for n>0 such that n>N.
(c) Show that limnlognn=0.

Since ddx (logxx) = ddx(x-12logx) = x-121x+ -12x-32 logx = x-32 (1-12logx) = 12x-32 (2-logx) is less than 0 for x>e2, is equal to 0 for x=e2, and is greater than 0 for x<e2, the function f(x)=log(x)|x12| is decreasing for x>e2 and has a maximum at e2.

Since log(x)0 for x1 and |x12|>0 for x1, an=logn|n12|0 for n>0 1 e 2 x 2 e y So an is bounded by 2e and 0 (0an2e) and an is decreasing if n>e2 (in particular, if n>9).

(c) limn lognn12 = limn logn(elogn)12 =limn logne12logn = limy ye12y= limy y1+12y+12!(12y)2+ = limy 11y+12+121122y+13!123y2+ limy 112!122y =limy8y=0.

Analyse the sequence an=(-1)n-1(1+1n). 1 2 3 4 5 6 7 n - 2 - 1 1 2 an is bounded above by 2,
an is bounded below by -2,
lim supan=1 and lim infan=-1,
supan=2 and infan=-32.

limnan does not exist because, as n gets larger and larger, an oscillates between close to lim supan (1, in this case) and lim infan (-1, in this case).

If an is a sequence in (or a totally ordered set X) such that

(a) an is increasing,
(b) an is bounded, and
(c) supan exists,
then (an) converges to supan.

Note: In , supan always exists, in , supan does not always exist.

Proof.

Let =supan. To show: limnan=.
To show: If ε>0 then there exists N>0 such that if n>0 and n>N then d(an,)<ε.

Proof by contradiction.
Assume that there exists ε>0 such that there does not exist N>0 such that if n>0 and n>N then d(an,)<ε. So, if N>0 then there exists n>0 with n>N such that d(an,)>ε.
so, if N>0 then -ε>anaN.
So -ε is an upper bound of (an).
Contradiction to =supan.
So limnan=.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100322suggLect10.pdf and was given on 22 March 2010.

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