By I. Craw

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We estimate the partial sums. Since 1/n2 > 0, clearly {Sn } is an increasing sequence. 9), it is convergent. From the diagram, 1 1 1 + 2 + ··· + 2 2 2 3 n Sn n dx , and so 2 1 x 1 n 1 < 1+ − ≤2− . x 1 n < Thus Sn < 2 for all n, the sequence of partial sums is bounded above, and the series is convergent. 8. Proposition. Let an be convergent. Then an → 0 as n → ∞. Proof. Write l = limn→∞ Sn , and recall from our work on limits of sequences that Sn−1 → l as n → ∞. Then an = (a1 + a2 + . . an ) − (a1 + a2 + .

Sin ≤ |an | → 0 as an an n → ∞. Infinite limits There are many more definitions and results about limits. 25. Definition. Say that limx→∞ f (x) = l iff given whenever x > K, then |f (x) − l| < . > 0, there is some K such that x2 + 3 . x→∞ 3x2 + 2x + 1 Solution. The idea here should be quite familiar from our sequence work. We use the fact that 1/x → 0 as x → ∞. 26. Example. Evaluate lim x2 + 3 1 1 + 3/x2 → = 2 2 3x + 2x + 1 3 + 2/x + 1/x 3 as x → ∞. 38 CHAPTER 4. 27. Definition. Say that limx→∞ f (x) = ∞ iff given L > 0, there is some K such that whenever x > K, then f (x) > L.

Say that limx→a− f (x) = l, or that f has a limit from the left iff given > 0, there is some δ > 0 such that whenever a − δ < x < a, then |f (x) − f (a)| < . 10. Example. Define f (x) as follows: 3 − x if x < 2; f (x) = 2 if x = 2; x/2 if x > 2. Calculate the left and right hand limits of f (x) at 2. 4. RESULTS GIVING CONINUITY 35 Solution. As x → 2−, f (x) = 3 − x → 1+, so the left hand limit is 1. As x → 2+, f (x) = x/2 → 1+, so the right hand limit is 1. Thus the left and right hand limits agree (and disagree with f (2), so f is not continuous at 2).

### Advanced Calculus And Analysis by I. Craw

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