8.2 Nonnegative Series 471If n = 2k + 1,
1 {l l/(2k+l) -k 1 1 1 (J3) 2k:+.l
~= 2k+yf,z = (3-k) = 32k+1 = 3 22k+1-2 = J3
1
-+-
J3
1 - 1 -
by Example 2.3.9. Thus, R = J3 and R = y12· Since R < 1, the root test
proves that the given series converges. D
*RAABE'S TEST
Several sophisticated tests for convergence of a nonnegative series I:: ak
have been developed for use when the ratio and root tests are inconclusive.
We include Raabe's test here because it is reasonably straightforward. Other,
more complicated, tests such as Kummer's test and Gauss' test can be found in
references such as [39], [47], [73], [85], [86], [87] [102], [111], [130], and [131]. To
understand where Raabe's test (and the others) come from it is helpful to look
at the situation in which the ratio test is inconclusive: that is, when ak+l ---+ 1.
ak
It is plausible to expect I:: ak to diverge if { a::^1 } converges to 1 "too rapidly,"
and I:: ak to converge if { a::^1 } converges to 1 "slowly enough." Raabe's test
captures this idea in mathematical precision.
Theorem 8.2.20 (Raabe's Test) Suppose I:: ak is a series of positive terms.
(a) If :3 r > 1 3 ak+l s; 1 - _kr for all but finitely many k, then L ak
ak
converges.
ak+l r.
(b) If :3 0 < r s; 1 3 2 1 - -k for all but finitely many k , then L ak
ak
diverges.Proof. (a) Suppose the hypotheses of (a) hold. Then :3 r > 1, no EN 3
ak+l r
k>no=>--<1--- ak - kThus, for n 2 no,=> kak+i s; kak - rak = (k - l)ak - (r - l)ak
=> (k - l)ak - kak+l 2 (r - l)ak > 0.n n
L [(k - l)ak - kak+1] 2 I:: (r - l)ak.
k=no k=no•An asterisk with a theorem, proof, or other material in this chapter indicates that the
material is challenging and can be omitted.