1. Let (an )n∈N , (bn )n∈N sequences, assume limn→∞ an = +∞ and bn is bounded below. Prove
that limn→∞ (an + bn ) = +∞.
Proof.
Let L be a lower bound for bn , that is, bn > L for all n ∈ N.
Fix K ∈ R. Since limn→∞ an = +∞, we can find an N such that an > K 0 := K − L for all
n > N.
Now, we will check this N works for an + bn . Indeed, for n > N we have
an + b n > K 0 + b n
= K + (bn − L)
> K
where in the last inequality we have used bn − L > 0.
Therefore limn→∞ (an + bn ) = +∞.
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2. Find the limit of the sequence an = (7 + 3n n) 3n .
Proof.
An educated guess tell us that for n large, the expression inside the parenthesis is like 3n n
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and hence the limit should be equal to lim(3n n) 3n = 3 3 lim n 3n = 3 3 · 1.
We prove the previous by using the sandwich rule, note first the following inequality
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(3n n) 3n < (7 + 3n n) 3n < (3 · 3n n) 3n
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Now by using the following known limits: lim 3 n = lim n n = 1, and product and power
rules we have
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lim(3n n) 3n = 3 3 lim n 3n
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= 3 3 (lim n n ) 3
1
= 33
and
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lim(3 · 3n n) 3n = 3 3 lim 3 3n lim n 3n
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= 3 3 (lim 3 n ) 3 (lim n n ) 3
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= 33
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Hence we conclude by sandwich rule, lim(7 + 3n n) 3n = 3 3 .
Finally a couple of limits for you to practice:
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• an = (n3 − 3n + 1) n .
√
• bn = n2 + 1 − n.
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