Chapter 8: Infinite Sequences and Series
Section 8.3: Convergence Tests
Determine if the series ∑n=0∞1n3+5 diverges, converges absolutely, or converges conditionally.
If it converges conditionally, determine if it also converge absolutely.
Since n3+5 behaves as n3/2 for large n, comparison to the convergent p-series Σ n−3/2 suggests the calculation
limn→∞1n3+5n−3/2 = limn→∞n3/2n3+5=1
from which, by part (1) of the Limit-Comparison test, the convergence of the given series is established. Since it is a convergent series with positive terms, the series converges absolutely.
As per the Mathematical Solution above, the absolute convergence of the given series can be established by the Limit-Comparison test, by a comparison to the convergent p-series ∑n=1∞1n3/2. Absolute convergence can also be established by the Integral test, but the integration is not elementary.
Figure 8.3.3(a) contains a graph of the function fx=1/x3+5 (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≥1. (The derivative appears to be negative for x>1.)
Consequently, the Integral test may be tried, provided the integration starts from, say, x=2. This is a nontrivial integration, one Maple evaluates in terms of the special function EllipticF.
To this end:
Calculus palette: Definite integral template
Context Panel: Evaluate and Display Inline
Context Panel; Approximate≻5 (digits)
Figure 8.3.3(a) Graph of fx (red) and f′x (green)
∫2∞1x3+5 ⅆx = EllipticF⁡2⁢3⁢51/3⁢2+51/32+51/3+3⁢51/3,14⁢2⁢3+14⁢23⁢51/3→at 5 digits1.3635
Since the integral converges, the series converges absolutely (as it is a series of positivel terms).
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