Calc II Exam 2
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Last updated 12:22 AM on 3/13/23
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9 Terms
1
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Geometric Series Test
if r >= 1 - the series diverges
if r < 1 - the series = c/(1-r) and converges
if r < 1 - the series = c/(1-r) and converges
2
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Integral Test
if f(x) is positive and decreasing - series converges if integral converges
if f(x) is positive and decreasing - series diverges if integral diverges
if f(x) is positive and decreasing - series diverges if integral diverges
3
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Divergence Test
if lim a(n) != 0 - series diverges
if lim a(n) = 0 - series converges
if lim a(n) = 0 - series converges
4
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Direct Comparison Test
if a(n)
5
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Limit Comparison Test
if a(n) and b(n) are positive - L = lim a(n)/b(n)
if not L>= 0 or L = inf - if a(n) converges then b(n) converges
if not L>= 0 or L = inf - if a(n) diverges then b(n) diverges
if L != 0 or inf - use another test
if not L>= 0 or L = inf - if a(n) converges then b(n) converges
if not L>= 0 or L = inf - if a(n) diverges then b(n) diverges
if L != 0 or inf - use another test
6
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Alternating Series Test
for series (-1)^n\*b(n)
if b(n) > 0 and b(n+1) > b(n) and lim b(n) = 0
then - series (-1)^n\*b(n) converges
if b(n) > 0 and b(n+1) > b(n) and lim b(n) = 0
then - series (-1)^n\*b(n) converges
7
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Ratio Test
L = lim |a(n+1)|/|a(n)|
if L < 1 - series a(n) converges
if L > 1 or lim = inf - series a(n) diverges
if L = 1 - try another test
if L < 1 - series a(n) converges
if L > 1 or lim = inf - series a(n) diverges
if L = 1 - try another test
8
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Root Test
L = lim nsqrt(|a(n)|)
if L < 1 - series a(n) converges
if L > 1 or lim = inf - series a(n) diverges
if L = 1 - try another test
if L < 1 - series a(n) converges
if L > 1 or lim = inf - series a(n) diverges
if L = 1 - try another test
9
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P - Test
If p > 1, then the series converges,
If p ≤ 1, then the series diverges.
If p ≤ 1, then the series diverges.
Note 
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