# Worksheet 11 MATH 3283W Fall 2012 Basic definitions. A sequence (s

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Worksheet 11 MATH 3283W Fall 2012 Basic definitions. A sequence (s
```MATH 3283W
Worksheet 11
Fall 2012
Basic definitions. A sequence (sn ) of real numbers is said to converge to s ∈ R if for any
> 0, there exists a number N such that for any n ≥ N the inequality |sn − s| < holds.
The number s is called the limit of sequence (sn ) and denoted by lim sn . If a sequence (sn )
does not converge to any number, it is said to diverge.
1. Prove the following statements using only the basic definitions.
a) The sequence (an ) defined by an =
n
n2 +1
converges to a = 0.
b) The sequence (bn ) defined by bn =
5n+1
7n−3
converges to b = 57 .
c) The sequence (cn ) defined by cn =
n3 −n
n3 +2n
d) The sequence (dn ) defined by dn =
√ n
n+100
e) The sequence (en ) defined by en =
converges to c = 1.
diverges.
√
√
n + 1 − n converges to d = 0.
f) The sequence (fn ) defined by fn = (−1)n diverges.
g) The sequence {0.2, 0.22, 0.222, 0.2222, . . . } converges to g = 92 .
h) The sequence (hn ) defined by hn = 0.99n converges to h = 0, but the sequence (jn )
with jn = 1.01n diverges.
2. Prove that if a sequence (sn ) converges to a strictly positive number s, then this
sequence contains infinitely many positive terms.
In practice, the following facts can be very useful.
Theorem.1 Let (sn ) and (tn ) be convergent sequences with lim sn = s and lim tn = t. Then
the following is true:
a) lim(sn ± tn ) = s ± t
b) lim(sn · tn ) = s · t
c) lim stnn = st provided that tn 6= 0 for all n ∈ N and t 6= 0.
d) for any k ∈ R, lim(k · sn ) = k · s and lim(k + sn ) = k + s
Lemma(commonly known as the squeeze lemma or two policemen lemma). Let (an ) and
(cn ) be convergent sequences such that lim an = lim cn = b. Then if (bn ) is such a sequence
that an ≤ bn ≤ cn for all n, then (bn ) converes and lim bn = b.
3. For each of the given sequences, find its limit (if it exists).
a)
b)
c)
d)
e)
1
1
+ 3n+5
.
an = 2n
2n−1
(n+3)10
bn = n10 +310 .
2)
√
.
cn = 10 sin(n
n
2n +10
dn = √
.
3n −10
n3√
−1+100
en = n n+5 .
Theorem 17.1, p.166.
2
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