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**Example text**

1) Is the series coarsely divergent? We have |an | = ln n →0 n+1 for n → ∞ by the law of magnitudes, so the series is not coarsely divergent. 2) Is the series absolutely convergent? Since ln n > 1 for n ≥ 3, we get the following estimate of the numerical series. ∞ ∞ ∞ 1 1 ln n ≥ = = ∞ (the harmonic series). n + 1 n + 1 n n=3 n=4 n=1 It follows from the criterion of comparison that the series is not absolutely convergent. DIVERSE - INNOVATIVE - INTERNATIONAL Please click the advert Are you considering a European business degree?

In case of convergence, check if the series is conditionally convergent or absolutely convergent. We have from the magnitudes, 0 < |an | = 3 · n 1 ≤ n n 3 2 for n ≥ n0 . The larger series 2−n is convergent, hence the smaller series is also convergent by the criterion of comparison, and the series is absolutely convergent. Alternatively the convergence is obtained by the criterion of roots, n |an | = √ n 3· √ 1 1 n n· → <1 3 3 for n → ∞. Alternatively the convergence is obtained by the criterion of quotients, |an+1 | 3(n + 1) 3n n+1 1 1 = = · → <1 · |an | 3n+1 3n n 3 3 for n → ∞.

1 It is possible directly to ﬁnd the sum. In fact, ∞ ϕ(x) := 1 = xn 1 − x n=0 for x ∈ ] − 1, 1[, hence by diﬀerentiation each term (which is legal for power series in their open interval of convergence), ∞ ϕ (x) = 1 = nxn−1 (1 − x)2 n=1 for x ∈ ] − 1, 1[. com 59 Calculus 3c-2 Conditional covergence and Leibniz’s criterion Choosing x = − 31 ∈ ] − 1, 1[, we get ∞ ∞ (−1)n−1 1 = n − n−1 3 3 n=1 n=1 n−1 =ϕ − 1 3 = 1 1+ 1 3 2 = 9 . 14 Check if the series ∞ (−1)n √ n−3 n n=10 is convergent or divergent.