epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems
2n-lp
EXAMPLE 4. Consider the sequence
Prove that
"+1 Jn=l
lim — = 2.
n->oo n + 1
Proq/; Let £>0 be given. Is it possible to have \2n-l <s for n
n+1
large enough?
(To get a feeUng for the behavior of the terms of the given sequence, one
might explicitly find some of them. Here is a very small collection of values:
ai = (2 - 1)/(1 + 1) = 0.5, a2 = (4- l)/(2 + 1) = 1, as = 1.25, aj = 1.625,
a49 = 1.94 as = 1.6, aig = 1.85, and a29 = 1.9. So, it does seem to be true
that as n gets larger the values of the terms of the sequence get closer and
closer to 2.)
Again, let us start by simplifying the expression that involves n, as shown
here:
2n-l
n + 1 -2 (2n - 1) - 2(n + 1)
n+1
-3
n+ 1
-31
|n+l|
n+r
Note that the number n 4-1 is positive, so it is equal to its absolute value.
Thus, one must choose n large enough to have:
3
n+ 1 < s.
This is equivalent to requiring:
n+1 1
99
or
n> — 1
s
Thus, let Ns = (3/6) - 1.
2n-l
For all n>Ne = (3/f) — 1 one will have
n+1
that 2 is indeed the limit of the sequence.
-2 < s, therefore proving
In the case of sequences as well, it is possible to check that the estimate
found for Ng is correct. Again, for some mathematicians this checking
process is an important part of the proof.
In Example 4, let n be any number such that n>iVe = (3/^) — 1, and let
\2n-l \
£>0 be given. Is it true that •-2 <sl Let us check. Consider the
n+1
equality:
|2n-l
n+1 -2 (2n - 1) - 2(n + 1)
n+1 n+1