Hence lim sup s n t n L M 2 \u03b5 since lim sup s n t n is the least upper bound of

# Hence lim sup s n t n l m 2 ε since lim sup s n t n

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Hence lim sup( s n + t n ) L + M + 2 ε , since lim sup( s n + t n ) is the least upper bound of the subsequential limits. Since this works for any ε > 0, we must have lim sup( s n + t n ) L + M, as desired. (b). Find an example to show that equality may not hold in part (a). Solution: Let s n = ( - 1) n and t n = ( - 1) n +1 . Then s n + t n = 0 for all n , so lim sup n →∞ ( s n + t n ) = lim n →∞ ( s n + t n ) = 0 . However lim sup s n = 1 and lim sup t n = 1. 5

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• Fall '07
• Allum
• Math, lim, Limit of a function, Limit of a sequence, Limit superior and limit inferior, Sn, subsequence