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Proof. By definition of limit, for every ɛ > 0 there exists nɛ ∈ N such that for every n ≥ nɛ we have: L − ɛ < an+1 − an bn+1 − bn < L + ɛ Since bn is increasing the number bn+1 − bn is always positive so we can multiply the inequality: (L − ɛ)(bn+1 − bn) < an+1 − an < (L + ɛ)(bn+1 − bn) If k ≥ nɛ, we can sum the three members of the inequality from nɛ up to k. Since the inequalities hold for each one of the summands, they also hold for the summation: (L − ɛ) k (bn+1 − bn) < n=nɛ k (an+1 − an) < (L + ɛ) n=nɛ k ((bn+1 − bn)) In each one of these summations, only the first and the last element don’t cancel out, so we get: n=nɛ (L − ɛ)(bk+1 − bnɛ) < ak+1 − anɛ < (L + ɛ)(bk+1 − bnɛ) If k is large enough we can assume that bk+1 is positive, since limk→+∞ bk = +∞, so we can divide everything by bk+1: (L − ɛ) 1 − bnɛ + bk+1 anɛ bk+1 < ak+1 bk+1 < (L + ɛ) 1 − bnɛ + bk+1 anɛ bk+1 The number nɛ is fixed, and bk+1 goes to +∞ when k is large; for this reason both anɛ converge to zero. We can find kɛ ∈ N such that for all k ≥ kɛ we have: and also: Using these last two inequalities we get: which proves that: Example 8.2. Calculate the limit: −ɛ < anɛ − (L + ɛ) bk+1 bnɛ bk+1 −ɛ < anɛ − (L − ɛ) bk+1 bnɛ bk+1 L − 2ɛ < ak+1 bk+1 ak+1 lim k→+∞ bk+1 < L + 2ɛ = L 1 lim n→+∞ k + 2k + . . . + nk nk+1 According to Stolz-Cesàro the limit is equal to the following: lim n→+∞ (n + 1) k (n + 1) k+1 = lim − nk+1 n→+∞ k+1 i=0 < ɛ < ɛ (n + 1) k ni − nk+1 k+1 i 10 = lim n→+∞ bk+1 (n + 1) k ni = k i=0 k+1 i and bnɛ bk+1
= lim n→+∞ (1 + n −1 ) k k i=0 k+1 i n i−k = 1 k+1 k = 1 k + 1 If k ≤ −1 the denominator is not an increasing sequence so the method cannot be applied. We will prove later on with some other technique that for k = −1 we have: Example 8.3. Calculate the limit: lim n→+∞ 1−1 + 2 −1 + . . . n −1 = +∞ lim n→+∞ n k=1 1 k − log n log n According to Stolz-Cesàro the limit is equal to the following: 1 1 n+1 − log (n + 1) + log n n+1 − log 1 + lim = lim n→+∞ log (n + 1) − log n n→+∞ 1 n = lim n→+∞ log 1 + 1 n 1 1+ 1 n − n log 1 + 1 n log 1 + 1 n n = 0 We don’t know if the numerator is convergent or not, but this calculation shows that it cannot diverge faster than log n. It also shows that n k=1 1 k goes to infinity as fast as log n. 11
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