notes3 - CM221A ANALYSIS I NOTES ON WEEK 3 UPPER AND LOWER...

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CM221A ANALYSIS I NOTES ON WEEK 3 UPPER AND LOWER LIMITS One says that + is an accumulation point of a sequence { a n } if there is a subse- quence of { a n } which converges to + . Similarly, -∞ is an accumulation point of { a n } if there is a subsequence which converges to -∞ . This convention makes sense, even though ±∞ are not proper numbers. Definition. The largest accumulation point of a sequence { a n } is called the upper limit of { a n } and is denoted lim sup a n . The smallest accumulation point of { a n } is called the lower limit of { a n } and is denoted lim inf a n . The Bolzano–Weierstrass Theorem implies that, for any bounded sequence, both upper and lower limits are finite numbers. If the sequence is unbounded the upper and lower limits may be ±∞ . Example. If a n = n then lim sup a n = lim inf a n = + . Note that the upper and lower limits always exist, even if the sequence does not converge. Exercise. Prove that a sequence { a n } has a limit (finite or infinite) if and only if lim sup a n = lim inf a n . Lemma. If r > 0 then lim sup( r a n ) = r lim sup a n and lim inf( r a n ) = r lim inf a n . If r < 0 then lim sup( r a n ) = r lim inf a n and lim inf( r a n ) = r lim sup a n . Proof. Let r 6 = 0. From the algebraic rules for limits it follows that a subsequence { a n k } converges to a limit a if and only if { r a n k } converges to the limit r a . There- fore c is an accumulation point of { a n } if and only if r c is an accumulation point of the sequence { r a n } . In other words, the accumulation points of the sequence { r a n } are obtained from the accumulation points of { a n } by multiplying them by the number r . If r > 0 then the multiplication maps the largest and smallest accumulation points of { a n } into the largest and smallest accumulation points of { r a n } . If r <
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notes3 - CM221A ANALYSIS I NOTES ON WEEK 3 UPPER AND LOWER...

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