Chapter 8: Infinite Sequences and Series
Section 8.3: Convergence Tests
Determine if the series ∑n=1∞−1n+12 n+5 diverges, converges absolutely, or converges conditionally.
If it converges conditionally, determine if it also converge absolutely.
Since the series is alternating, it is a candidate for the Leibniz test. Now an=1/2 n+5, which is monotone decreasing to zero as n→∞. Hence, by this test, the series converges conditionally.
However, the Limit-Comparison test, with the harmonic series as Σ bn, gives
c=limn→∞anbn = limn→∞1/2 n+51/n = limn→∞n2 n+5=1/2>0
Since c=1/2>0, and Σ bn (the harmonic series) diverges, it follows that Σ an also diverges. Hence, the given alternating series does not converge absolutely. It only converges conditionally.
Obtain the sum of the (alternating) series
Expression palette: Summation template
Context Panel: Evaluate and Display Inline
∑n=1∞−1n+12 n+5 = 1315−14⁢π
Finding the explicit sum of this alternating series would actually establish its conditional convergence, provided that Maple's internal manipulations were exposed. Short of that, either the details of the summation would need to be provided, or the Leibniz test implemented.
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