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This series cannot be made any simpler. We can try to apply the 2 k -test and we obtain: +∞ n=1 n α 2 n(1−β) We conclude by comparing with the improper integral: +∞ It becomes more familiar if we replace 2 with e: +∞ 1 1 x α 2 x(1−β) dx x α e x(1−β) ln(2) dx If β − 1 > 0 we can make the substitution t = x(β − 1) ln(2) and we obtain: +∞ x (β−1) ln(2) α e −x dx which is convergent for every value of α. The series is convergent for β > 1 and α > 0 and not convergent for 0 < β ≤ 1 and any value of α. We could have skipped the 2 k -test and apply the integral test directly (this second approach is not any simpler). 5 Fundamental series The following series should be considered fundamental, and you don’t have to retrieve them each time you do a calculation: • • • • +∞ n=1 It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test. It is convergent if and only if |p| < 1. It can be calculated explicitly. +∞ n=1 +∞ n=1 1 n p p n 1 n(ln(n)) p It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test. +∞ n=1 It is convergent. It can be done by comparison. 4 n! n n
• • • It is convergent. Done by comparison. It is convergent for every p. It can be retrieved with the integral test. +∞ n=1 +∞ n=1 +∞ n=1 e n n! n p e n ln(n) n p It is convergent if and only if p > 1. It can be retrieved with the integral test or the 2 k -test. The number p in all the previous formulas is a real number. 6 Good but not optimal estimate of the factorial Sometimes it’s useful to compare the factorial with some better looking function. We start from the identity: n ln(n!) = ln(i) Since ln is monotonic (increasing) we can compare with the integral: n 1 ln(x) dx ≤ i=1 n ln(i) ≤ i=1 n+1 1 ln(x) dx We can integrate the logarithm and we have: n n ln(n) − n + 1 ≤ ln(i) ≤ (n + 1) ln(n + 1) − n We can now remove the logarithm by exponentiating the inequality: i=1 nn (n + 1)n+1 ≤ n! ≤ en−1 en Example 6.1. Determine if the series is convergent or divergent: Applying the estimate we have: Since the series +∞ n=1 well. +∞ n=1 3 n n! n n 3n en−1 ≤ 3nn! nn ≤ 3n (n + 1) n+1 nnen n 3 e is divergent, by comparison the series we are studying is divergent as 5
Page 1 and 2: 1 Definitions and properties Series
Page 3: 4 Strategy There is no algorithm to

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