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Nth-term test for DIVERGENCE
always do first
diverge if limit ā 0
Geometric series test
if r < 1 converge
if r >= 1 diverge
find S = a/1-r
Ratio test DO ENDPOINTS!!!
find An+1
limit ABOLUTE VALUE then DIVIDE An+1 by An
if limit < 1, series converges
if limit > 1, series diverges
if limit = 1, test is inconclusive
write everything out, declare terms
Integral test
f(x) MUST BE
POSITIVE, DECREASING, CONTINUOUS
find the integral of bound
if the limit of integral ā infinity converges
if the limit of integral = infinity diverges
P-series test
for pure 1/n
p > 1 converges
p ā¤ 1 diverges
Direct Comparison Test
POSITIVE
mutated series, but not as much
SIMPLIFY, find CONVERGENCE
big converge small converge >
small diverge big diverge <
Limit Comparison Test
POSITIVE
Limit, Longer Test + Terms
SIMPLIFY, find CONVERGENCE
LIMIT of NORMAL / SIMPLIFIED =
if it approaches FINITE number, both converge or diverge
Alternating Series Test for CONVERGENCE
MUST BE:
-alternating
-decrease in magnitude (compare An > An+1)
-have limit of 0
Then, REMOVE the alternator, take the limit, and use nth term test
If conditions match, series converge
Absolute and Conditional Convergence
Check convergence for abs value and normal
-if abs value of An converge, absolute convergence
-if An converge but abs value of An donāt, conditional convergence
integral tan inverse formula
1/a tan-1(u/a) + c
integral sin inverse
sin -1 (u/a) +c
Alternating Series Remainder
expand
must write: since series is
decreasing in magnitude
alternating
limit of zero
Sn - error <= Sn <= Sn + error
error is the absolute value of the next term
show that error < fraction number
Alternating series ERROR
WRITE CONDITIONS FOR AST
alternating, decrease, limit of 0
find term that is less than error
state that S - abs error <= Sum <= S + abs error
declare all terms