# This is a bit harder we assume that relation 1 is

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( ) This is a bit harder. We assume that relation (1) is true and we’ll construct inductively a subsequence { x n k } k which will approach as k → ∞ . To pick the first term of our subsequence, apply first (1) with M = M 1 = 1. It follows that there exists n 1 N such that x n 1 > 1. Now the idea is to apply (1) with a bigger and bigger M . But just choosing M = 2 , M = 3, etc. may not work, because we have no guarantee that the ranks n 2 , n 3 , etc. which we get from (1) will be in increasing order. Here is how we pick the second term of the subsequence. Let M 2 = max { 2 , x 1 , x 2 , ..., x n 1 } and apply (1) with M = M 2 . It follows that there exists n 2 N such that x n 2 > M 2 . From the choice of M 2 we deduce two things. First, x n 2 > M 2 2. Secondly, x n 2 > M 2 x l , for any l ∈ { 1 , 2 , ..., n 1 } ; in particular, we get that n 2 > n 1 because otherwise x n 2 would be equal to one of the x l , l ∈ { 1 , 2 , ..., n 1 } . Suppose now that we picked n 1 < n 2 < ... < n k , such that x n l > l , for any l ∈ { 1 , 2 , ..., k } and we’ll construct the k + 1-th term of the subsequence. Let M k +1 = max { 2 , x 1 , x 2 , ..., x n k } and apply (1) with M = M k +1 . It follows that there exists n k +1 N such that x n k +1 > M k +1 . From the choice of M k +1 it follows as above that n k +1 > n k and that x n k +1 > k + 1. Thus by induction we construct the subsequence { x n k } k of { x n } n , with the property that x n k > k , for any k N . The (extended) comparison Theorem implies immediately that x n k → ∞ as k → ∞ .
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