27 Jul 2016

School

Department

Course

Professor

Economics 211

Some notes on chapters 3 and 4

The text deﬁnes a sequence as a function whose domain is the natural numbers {1,2,3, . . .}

(page 62). So if the range is <nwe could write the sequence as a countable subset of <n.

{a1, a2, a3, . . .}≡{ai}∞

i=1 ⊂ <n.

The limit of {ai}∞

i=1 as i→ ∞ exists and is a∈ <nif for each > 0 there is some K

(which depends on ) in the natural numbers such that

{ai}∞

i=K⊂N(a).

For example, {2−1

i,3 + 1

i}∞

i=1 is a sequence in <2whose limit is (2,3).

Aseries, say {si}∞

i=1, is a sequence that is based another sequence, say {ai}∞

i=1 ⊂ < (not

<n, n ≥2 ), with the structure that s1=a1,s2=a1+a2,s3=a1+a2+a3, and so on. Will

the limit of {si}∞

i=1 exist? If ai>0 for each iand

for i= 1,2, . . . ai+1

ai

≤rwhere 0 < r < 1,

then the answer is yes. Consider the geometric series deﬁned by

gi=Ar0+Ar1+. . . +Ari−1for i= 1,2,....

Oberve that

rgi=Ar1+. . . +Ari−1+Ari

Subtracting the last two equations

(1 −r)gi=Ar0−Ari=A1−ri

or

gi=A1−ri

1−r

If r < 1rigoes to zero and gigoes to A/(1 −r) as igoes to inﬁnity. So the geometric

series converges. If we can show this geometric series acts as an upper bound on {si}∞

i=1

under the conditions stated above then simust converge. Note that s1=a1,s2=a1+a2≤

a1+ra1=a1(1 + r), s3≤a1(1 + r+r2), and so on. Thus {si}∞

i=1 converges to some positive

number because it is bounded above by a geometric series that converges to a1/(1−r)∈ <+.

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