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100 The Nuts and Bolts of Proof, Third Edition
Because n > (3/6:) - 1, it follows that n + 1 > 0/s) > 0. Therefore,
\ s 3
< -. This inequality is equivalent to the inequality <£. So, we
n can + 1conclude 3 that i^v._ii 2n-l -2 <e whenever n>Ne = (3/s)
^
—
+ 1
1.
n+1
One thing to be noticed in the definition of limit is that while e must
always be a positive number, the number Ng can be negative, and, while n
is always an integer number, Ng does not need to be integer. Some authors
do specify that iVg should be a nonnegative integer, but there is no need
to do so.
If Ns happens to be negative, the statement n>Ns is always true. Thus,
the inequality |a„ - L\<e will be true for all values of n.
If Ne is not an integer, then n will just be any integer larger than it.
In Example 4, when ^ = 4, N4 = -1/4. This means that the statement
? ^ - 2I <4 is always Tu.. I, is easy .o check .hat to is .ndeed .he case
by looking at the values of the terms of the sequence < a„
When s = 4/5, N4/5 = 2.75. This means that the statement
2n-l , 4
-2 < - is true for all numbers n larger than 2.75. As n is an integer.
n+1
it means that it will be true for n > 3.
3n
rx.1 ,. 3n 3
EXAMPLE 5. Let \ an = . Then lim = -.
2n-l n^oo 2n— I 2
n=l
3n
Proof: Let £>0 be given. Is it possible to have
<s for n
large enough?
2n-l
We will start by simplifying the expression in the absolute value:
3n
2n-l
2 X 3n - 3(2n - 1)
2{2n - 1) 2(2n - 1) 2(2n - 1)
Therefore, one needs to solve the inequality:
3
2{2n - 1) ^ ^'
Doing so yields the result w> H4 + !)• ^^^^^ ^^^ ^^ = 2 (i + !)•
In the case of limits of sequences, similarly to the case of limits of
functions, given an ^ > 0 the corresponding Ng is not unique. See Exercises 9
and 10 at the end of this section.
For sequences, as is the case with functions as well, it is usually easier to
prove the existence of a limit than its nonexistence. Indeed, to prove
•