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

150 The Nuts and Bolts of Proof, Third Edition
7. To prove that lim„
>«2+l
= 0 means proving that for every given s>0
there exists an N such that ^^ —0 <£ for all n>N. Because
^^, we need to find values of n that satisfy the
3^-0
n2+l
1
«2 + l
inequality ((l/(n^ + 1))<^. This yields n^ -\-l>{l/s), or n^>(l/e) - 1.
This is a second-degree inequality, and its solution set is
n< — \l\le — 1 or n>'s/\/s — 1. Because n > 1, we will only choose
n>N = ^XJE - 1. Note that A/^ is a real number only when 1 > e>0.
In any case, when e>\, the inequality
n2+l
0 <E will be true for
all values of n. Indeed (l/(n^ + !))< 1 for all values of n.
8. To prove that lim„^oo f^ = \ means proving that for every given e > 0
there exists an N such that ||^ -1| <£ for all n>iV. Simplification of
the difference between the terms of the series and 5/3 yields
|5n+l
l3«-2'
3(5n+l)-5(3w-2)
3(3n-2)
|3(3n-2)
_ 13
~ 313«-2r
Because n > 1, the
expression 3n-2>0. Thus, ||^-3| — 3(3^^,2)' i^—, and we need to find
values of n that satisfy the inequality (13/(3(3n - T)))<E. Therefore, we
obtain n>A/^ ^ ^(^ +2).
9. Consider again the calculation performed in Example 4:
ln-\
{In - 1) - l(n + 1)
n+ 1 •-2 n-V 1
-3
n+ 1
1-31 ^ 3
|n+l| n+r
If n>M4/5 = 16, then n + 1 > 17; therefore, | ^ - 2| = ^ifj < ^ < 3.
10. Let us start by observing that
n + 1
0
n+1 1 1 1/ 1\
—y- = - + ^ = - 1+- .
Case 1. n>Ns = ^fl + Vl + 4g). Because n>N^= ^(1 + Vl + 4e),
((l/n)<(2£)/(l + V1 + 4^)). Moreover note that
(1 + VTT4^) = 1 + 2vTT4^ + 1 + 46: = 2 + 4£ + 2Vl + 4s.
Therefore, |^ - 0^ i (1 + ^) < T^fe (^ + ITTfe^)' T^^^'
n+1 26 /l + vTH^ + 2g\
l^-^/^T4^V •V i + yiT4^ /
•-(2 + 4^ + VTT4£) = £,
(i + yiT4^'2