Chapter 8: Infinite Sequences and Series
Section 8.3: Convergence Tests
Example 8.3.14
Determine if the series ∑n=2∞−1nlnn diverges, converges absolutely, or converges conditionally.
If it converges conditionally, determine if it also converge absolutely.
Solution
Mathematical Solution
The series is alternating, suggesting the Leibniz test. For this, 1/ lnn must eventually become a monotone decreasing sequence with limit zero. Consequently, the following calculations are executed.
limn→∞1/ lnn = 0 and ddx1/ lnx=−1x lnx2⁢<0 for x>1
The conditions of the Leibniz test apply, and by that test, the series converges conditionally.
To test for absolute convergence, apply the Integral test: ∫2∞1lnx ⅆx = ∞.
By this test, the series of absolute values does not converge, so the series is not absolutely convergent.
Maple Solution
Figure 8.3.14(a) contains a graph of the function fx=1/lnx (in red) and of its derivative (in green).
On the basis of this graph, it may be conjectured that f is monotone decreasing and bounded below by zero, provided x≥2. (The derivative appears to be negative for x>1.)
Consequently, the Integral test may be tried as a test for absolute convergence, provided the integration starts from, say, x=3.
To this end:
Calculus palette: Definite integral template Context Panel: Evaluate and Display Inline
∫2∞1lnx ⅆx = ∞
module() local F,p,N; F:=1/ln(x); N:=10; p:=plot([F,diff(F,x)],x=1..N,color=[red,green],view=[0..N,-5..5],tickmarks=[N,4],labels=[x,y]); print(p); end module:
Figure 8.3.14(a) Graph of fx (red) and f′x (green)
Since the integral diverges, the series does not converge absolutely.
The following two calculations support the use of the Integral test.
Calculus palette: Limit operator
Context Panel: Evaluate and Display Inline
limn→∞1lnn = 0
Calculus palette: Differentiation operator
ⅆⅆ x 1lnx = −1ln⁡x2⁢x
The derivative is negative for x>1. (This is consistent with Figure 8.3.14(a).) Since the criteria of the test are satisfied, it follows that the series does not converge absolutely. The Leibniz test applies, and by it, this alternating series is seen to converge conditionally.
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