New series whose terms are severally greater than

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new series whose terms are severally greater than those of the divergent series δ + 1 2 δ + 1 2 δ + . . . ; and therefore u n is divergent. Examples LXIX. 1. Use Abel’s theorem to show that (1 /n ) and { 1 / ( an + b ) } are divergent. [Here nu n 1 or nu n 1 /a .] 2. Show that Abel’s theorem is not true if we omit the condition that u n decreases as n increases. [The series 1 + 1 2 2 + 1 3 2 + 1 4 + 1 5 2 + 1 6 2 + 1 7 2 + 1 8 2 + 1 9 + 1 10 2 + . . . , * This theorem was discovered by Abel but forgotten, and rediscovered by Pring- sheim.
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[VIII : 174] THE CONVERGENCE OF INFINITE SERIES, ETC. 393 in which u n = 1 /n or 1 /n 2 , according as n is or is not a perfect square, is convergent, since it may be rearranged in the form 1 2 2 + 1 3 2 + 1 5 2 + 1 6 2 + 1 7 2 + 1 8 2 + 1 10 2 + · · · + 1 + 1 4 + 1 9 + . . . , and each of these series is convergent. But, since nu n = 1 whenever n is a perfect square, it is clearly not true that nu n 0.] 3. The converse of Abel’s theorem is not true , i.e. it is not true that, if u n decreases with n and lim nu n = 0, then u n is convergent. [Take the series (1 /n ) and multiply the first term by 1, the second by 1 2 , the next two by 1 3 , the next four by 1 4 , the next eight by 1 5 , and so on. On grouping in brackets the terms of the new series thus formed we obtain 1 + 1 2 · 1 2 + 1 3 ( 1 3 + 1 4 ) + 1 4 ( 1 5 + 1 6 + 1 7 + 1 8 ) + . . . ; and this series is divergent, since its terms are greater than those of 1 + 1 2 · 1 2 + 1 3 · 1 2 + 1 4 · 1 2 + . . . , which is divergent. But it is easy to see that the terms of the series 1 + 1 2 · 1 2 + 1 3 · 1 3 + 1 3 · 1 4 + 1 4 · 1 5 + 1 4 · 1 6 + . . . satisfy the condition that nu n 0. In fact nu n = 1 if 2 ν - 2 < n 5 2 ν - 1 , and ν → ∞ as n → ∞ .] 174. Maclaurin’s (or Cauchy’s) Integral Test. * If u n decreases steadily as n increases, we can write u n = φ ( n ) and suppose that φ ( n ) is the value assumed, when x = n , by a continuous and steadily decreasing function φ ( x ) of the continuous variable x . Then, If ν is any positive integer, we have φ ( ν - 1) = φ ( x ) = φ ( ν ) when ν - 1 5 x 5 ν . Let v ν = φ ( ν - 1) - Z ν ν - 1 φ ( x ) dx = Z ν ν - 1 { φ ( ν - 1) - φ ( x ) } dx, * The test was discovered by Maclaurin and rediscovered by Cauchy, to whom it is usually attributed.
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[VIII : 174] THE CONVERGENCE OF INFINITE SERIES, ETC. 394 so that 0 5 v ν 5 φ ( ν - 1) - φ ( ν ) . Then v ν is a series of positive terms, and v 2 + v 3 + · · · + v n 5 φ (1) - φ ( n ) 5 φ (1) . Hence v ν is convergent, and so v 2 + v 3 + · · · + v n or n - 1 X 1 φ ( ν ) - Z n 1 φ ( x ) dx tends to a positive limit as n → ∞ . Let us write Φ( ξ ) = Z ξ 1 φ ( x ) dx, so that Φ( ξ ) is a continuous and steadily increasing function of ξ . Then u 1 + u 2 + · · · + u n - 1 - Φ( n ) tends to a positive limit, not greater than φ (1), as n → ∞ . Hence u ν is convergent or divergent according as Φ( n ) tends to a limit or to infin- ity as n → ∞ , and therefore, since Φ( n ) increases steadily, according as Φ( ξ ) tends to a limit or to infinity as ξ → ∞ . Hence if φ ( x ) is a function of x which is positive and continuous for all values of x greater than unity, and decreases steadily as x increases, then the series φ (1) + φ (2) + . . .
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