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lim inf(an+bn)=lim inf(an)+lim inf(bn) provided that liman exists

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Question. Suppose (an) and (bn) are two sequences of real numbers such that limnan=a. Show that lim infn(an+bn)=lim infn(an)+lim infn(bn). To avoid tedious discussion with infinity, assume in addition that both an and bn are bounded sequences.

Notes. The question is based on the Wikipedia article on limit inferior and limit superior, where the statement above is mentioned explicitly in the article without any proof.

It follows immediately from the definition of lim inf that lim infn(an+bn)lim infn(an)+lim infn(bn) (See this section of the article.)

Thus, the question reduces to showing the reversed inequality: lim infn(an+bn)lim infn(an)+lim infn(bn)

Based on a few basic observations of the definition of liminf, there is one "algebraic" way to show this inequality, which conceals the use of many quantifiers. I will write my own answer below. Alternatives are welcome.

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We will use the following facts in the proof:

  • (1) If the limit of (xn) exists, so does that of (xn)

  • (2) If the limit of xn as n exists, then lim infnxn=lim supnxn=limnxn

  • (3) One obvious direction of the inequality is proved by definition: lim infn(an+bn)lim infn(an)+lim infn(bn)

Proof.

lim infnbnlim infn(an+bn)+lim infn(an)by fact (3)=lim infn(an+bn)+limn(an)by fact (1,2)=lim infn(an+bn)limn(an)linearity=lim infn(an+bn)lim infn(an)by fact (2)

which implies that lim infn(an+bn)lim infn(an)+lim infn(bn)

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