By Schloemilch O.

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Constructed in this way is increasing. On the other hand, if C is unbounded above, we can find an infinite sequence of natural numbers nl < n2 < n3 < ... all of which lie in C. But by definition of C, ni > nl 26 Analysis means that x n 2 ~ X n l , n3 > n2 implies x n 3 this way is decreasing. ~ x n 2 ' etc. So the subsequence found in • Theorem 2-Bolzano-Weierstrass Theorem - - - Every bounded monotone sequence in IR has a convergent subsequence. PROOF Given the sequence (x n ) choose a monotone subsequence (x n, ) .

To deal with the variety of possible limits of subsequences of divergent sequences we introduce a new concept and some notation: • Definition 2 (i) A real number a is an accumulation point of the sequence (x n ) if some subsequence of (x n ) converges to a. (ii) Suppose (x n ) is a given real sequence. If (x n ) converges, then x = limn~oo x; is its only accumulation point, since for each e > 0 there are at most finitely many n for which IXn - x] 2: e. We shall write N(x, e) for the open interval (x - e, x + e) of radius e centred at x (we call this the e-neighbourhood of x), and say that nearly all elements of (x n ) belong to N(x, e) if the above statement is true.

Clearly, if r ::s 0 the terms ~ do not converge to 0, so the series must diverge. For r > 0 we can use the Condensation Test, which means we need to look at the geometric series Lk 2k(2 = Lk(2 1- r)k. This series converges if and only if2 1- r < 1, and this occurs exactly ifr > 1. Hence we have proved the series Ln ~ converges if and only if r > 1. Note how this distinguishes between the two special cases (r = 1 or 2) which we discussed earlier. t) Example 9 Of the above tests, the Ratio Test is much the easiest one to use.

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