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For all N∈N, define vN=sup{sn:n≥N} and uN=inf{sn:n≥N}; so by definition

lim supan=limN→∞vN and lim infan=limN→∞uN.

1) Suppose lim supan=lim infan=L, and let ϵ>0 be given.

a) Since lim supan=limN→∞vN=L,vN<L+ϵ for some N∈N so an<L+ϵ for n≥N.

b) Since lim infan=limN→∞uN=L,L−ϵ<uM for some M∈N so L−ϵ<an for n≥M.

If K=max{N,M}, then L−ϵ<an<L+ϵ for n≥K; so limn→∞an=L.

2) Suppose limn→∞an=L, and let ϵ>0 be given.

Then there is an N∈N such that |an−L|<ϵ for n≥N, so L−ϵ<an<L+ϵ for n≥N.

Therefore L−ϵ≤uN and vN≤L+ϵ, so L−ϵ≤lim infan and lim supan≤L+ϵ

since (uN) is increasing and (vN) is decreasing.

Since ϵ>0 was arbitrary, L≤lim infan≤lim supan≤L so lim infan=lim supan=L.

Explanation:

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