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# Tutorial-1 - Analysis II Tutorial Solution sketches...

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Analysis II Tutorial Solution sketches Alexandre Tomberg January 13, 2011 1 Let ( x n ) be a bounded sequence and let L be the set of its limit points. Let ( a k ) be a sequence in L and suppose that a = lim k →∞ a k exists. Then a L . Proof. To show a L , it is enough to find a subsequence ( x n j ) of ( x n ), s.t. lim j →∞ x n j = a. As a k L , subsequence ( x n ( k ) j ), s.t. a k = lim j →∞ x n ( k ) j . Thus, for each k , we an find an x n k s.t. | x n k - a k | < 1 k and n k > n k - 1 . Then, as | x n k - a | = | x n k - a k + a k - a | ≤ | x n k - a k | + | a k - a | < 1 k + | a k - a | , taking k → ∞ , we get that lim k →∞ x n k = a . 1

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2 Find the limsup and lininf of the following sequence x n = n 2 1 + n 2 cos 2 πn 3 . Solution. As n , - 1 2 = cos (2 π/ 3) = cos 2 π 3 k - 2 3 = cos 2 π 3 k - 1 3 < cos(2 πk ) = 1 . Clearly, the subsequences { x 3 k - 2 , x 3 k - 1 } and { x 3 k } converge, so that we get lim inf n →∞ x n = ! lim k →∞ x 3 k - 2 = lim k →∞ (3 k - 2) 2 1 + (3 k - 2) 2 · - 1 2 = - 1 2 , lim sup n →∞ x n = lim k →∞ x 3 k = lim k →∞ (3 k ) 2 1 + (3 k ) 2 = 1 . Note that all other convergent subsequences must be contained in the two subsequences we considered. 3 Construct a real sequence ( x n ), whose set of L of limit points is s.t.
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Tutorial-1 - Analysis II Tutorial Solution sketches...

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