Since a 1 a n and x n is the inf of a n a 1 x n Since x n a n a n 1 by

# Since a 1 a n and x n is the inf of a n a 1 x n since

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Since a - 1 < a n and x n is the inf of a n , a - 1 x n Since x n a n < a n + 1 by transitivity x n < 1 Therefore | x n - a | ≤ 1 = 2 < So by transitivity | x n - a | < Since wasarbitrary > o, N N such that n N, | y n - a | < and | x n - a | < Therefore y n and x n converge to a by definition of converges. So if lim a n exists then lim inf a n = lim sup a n Suppose lim inf a n = lim sup a n Since x n a n y n , by the Squeeze Theorem lim x n = lim a n = lim y n Therefore if lim inf a n = lim sup a n then lim a n exists So we have if lim inf a n = lim sup a n then lim a n exists and if lim a n exists then lim inf a n = lim sup a n Therefore lim inf a n = lim sup a n iff lim a n exists 5
4 Find an explicit formula for the sequence and determine if the series converges X n =1 1 n ( n + 1) X n =1 1 n ( n +1) = X n =1 1 n - 1 n +1 X n =1 1 n - 1 n +1 = 1 - 1 2 + 1 2 - 1 3 + 1 3 ... - 1 k + 1 k - 1 k +1 1 - 1 2 + 1 2 - 1 3 + 1 3 ... - 1 k + 1 k - 1 k +1 = 1 - 1 k +1 lim k →∞ 1 - 1 k +1 = 1 Therefore the series converges to 1 6

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• Fall '19
• lim, Mathematical analysis, Limit of a sequence