Convergence Tests Flashcards
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — DIVERGENCE TEST —
SERIES FORM: sum(ak)
CONVERGENCE: does not apply
DIVERGENCE: lim ak ≠ 0
COMMENTS: Cannot be used to prove convergence
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — GEOMETRIC SERIES —
SERIES FORM: ∑ark
CONVERGENCE: |r| < 1
DIVERGENCE: |r| ≥ 1
COMMENTS: If |r| < 1, then ∑ark = a/(1-r)
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — INTEGRAL TEST —
SERIES FORM: ∑ak, where ak=f(k) and f is continuous, positive, and decreasing
CONVERGENCE: ∫ f(x)dx < ∞
DIVERGENCE: ∫ f(x)dx does not exist
COMMENTS: The value of the integral is not the value of the series.
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — P-SERIES —
SERIES FORM: ∑1/kp
CONVERGENCE: p > 1
DIVERGENCE: p ≤ 1
COMMENTS: Useful for comparison tests
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — RATIO TEST —
SERIES FORM: ∑ak where ak > 0
CONVERGENCE: lim (ak+1)/(ak) < 1
DIVERGENCE: lim (ak+1)/(ak) > 1
COMMENTS: Inconclusive if lim (ak+1)/(ak) = 1
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — ROOT TEST —
SERIES FORM: ∑ak where ak ≥ 0
CONVERGENCE: lim k√(ak) < 1
DIVERGENCE: lim k√(ak) > 1
COMMENTS: Inconclusive if lim k√(ak) = 1
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — DIRECT COMPARISON TEST —
SERIES FORM: ∑ak, where ak > 0
CONVERGENCE: 0 < ak ≤ bk, and ∑bk converges
DIVERGENCE: 0 < bk ≤ ak, and ∑bk diverges
COMMENTS: ∑ak is given; you supply ∑bk
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — LIMIT COMPARISON TEST —
SERIES FORM: ∑ak, where ak > 0, bk > 0
CONVERGENCE: 0 < lim (ak/bk) < ∞, and ∑bk converges
DIVERGENCE: lim (ak/bk) and ∑bk diverges
COMMENTS: ∑ak is given; you supply ∑bk
For the given test, give the form of the series and list the condition of convergence, divergence, and comments. — ALTERNATING SERIES TEST —
SERIES FORM: ∑(-1)kak, where ak > 0, 0 < ak+1 ≤ ak
CONVERGENCE: lim ak = 0
DIVERGENCE: lim ak ≠ 0
COMMENTS: Remainder Rn satisfies Rn < an+1