How do you determine convergence/divergence of ∑n=1∞n21?
Identify p = 2. Since 2 > 1, the series converges.
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How do you determine convergence/divergence of ∑n=1∞n21?
Identify p = 2. Since 2 > 1, the series converges.
How do you determine convergence/divergence of ∑n=1∞n1?
Rewrite as ∑n=1∞n1/21. Identify p = 1/2. Since 1/2 < 1, the series diverges.
How do you determine convergence/divergence of ∑n=1∞n3n2?
Simplify to ∑n=1∞n1. Identify p = 1. Since p=1, the series diverges (Harmonic Series).
Determine convergence/divergence of ∑n=1∞n0.751
Identify p = 0.75. Since 0.75 < 1, the series diverges.
Determine convergence/divergence of ∑n=1∞n−4
Rewrite as ∑n=1∞n41. Identify p = 4. Since 4 > 1, the series converges.
Determine convergence/divergence of ∑n=1∞n5n
Simplify to ∑n=1∞n3/21. Identify p = 3/2. Since 3/2 > 1, the series converges.
Determine convergence/divergence of ∑n=1∞(n1)3
Rewrite as ∑n=1∞n31. Identify p = 3. Since 3 > 1, the series converges.
Explain the convergence/divergence of a p-series.
A p-series ∑n=1∞np1 converges if p > 1 and diverges if p ≤ 1.
Explain the behavior of the harmonic series.
The harmonic series ∑n=1∞n1 diverges, even though its terms approach zero.
Why is simplifying the terms important before applying the p-series test?
Simplification ensures the series is in the standard np1 form, allowing for correct identification of 'p' and accurate convergence/divergence determination.