By Even S.

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**Example text**

To derive this result, assume, ﬁrst, that a = 0 and b = 1. Divide the interval I = [a, b] = [0, 1] into 10 subintervals (of the same length), and call them I0 , . . 1). , {n : xn ∈ Id1 } is inﬁnite, and pick one of the terms of the sequence in Id1 and call it x1 . Then, the length of Id1 is 1/10. Now, divide Id1 into 10 subintervals again ordered left to right and called Id1 0 , . . , Id1 9 . Pick one of them, say Id1 d2 , containing inﬁnitely many terms of the sequence, and pick one of the terms (beyond x1 ) in the sequence in Id1 d2 and call it x2 .

As an application of this theorem, lim n 2 + n12 2n2 + 1 2+0 = lim = 2. 3). Now, let a > 1 and set b = a − 1. Then, ∞. 6). Hence, an 0. Summarizing, 0 < a < 1, then, a = 1/b with b > 1, so, an = 1/bn A. if a > 1, then, an ∞, B. if a = 1, then, an = 1 for all n ≥ 1, and C. if 0 ≤ a < 1, then, an → 0. Sometimes we say that a sequence (an ) converges to L if an → L. If the speciﬁc limit is not relevant, we say that the sequence converges or is convergent. If a sequence has no limit, we say it diverges.

If a = −1, sn equals 0 or 1 (depending on n), hence, diverges, hence, the series diverges. We have shown ∞ an = n=0 1 , 1−a if |a| < 1, ∞ and n=0 an diverges if |a| ≥ 1. To study more general series, we need their arithmetic and ordering properties. 1. If an = L and bn = M , then, (an + bn ) = L + M . If bn = cL = c( an ). If an ≤ bn ≤ cn an = L, c ∈ R, and bn = can , then, and an = L = cn , then, bn = L. To see the ﬁrst property, if sn , tn , and rn denote the partial sums of an , bn , and cn , then, sn +tn equals the partial sum of (an +bn ).