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Calc 2 test 3

ch 10.3, 10.4, 11.1-11.6

What is the Geo Series? diverges for absolute value of (r) >= 1 Converges for absolute value of (r)<1 to a sum of a/(1-r)
What is the Test for Divergence if the limit as n-->infinity of (a_n)= Does Not Exist OR limit as n-->infinity of (a_n) does not = 0 then the the Sum of (a_n) from n=1 to infinity is divergent
Comparison Test if both the Sum of a_n and the Sum of b_n are (+0 term then (a_n)>(b_n)>0 OR (a_n)>=(b_n)>=0 1) if the Sum of (a_n) converges then the Sum of (b_n) converges 2)if the Sum of (b_n) diverges then the Sum of (a_n) diverges
slowest to quickest terms going to infinity ln(n), n^a, a^n, n!, n^n
Alternating Series Test to use series MUST alternate every term alternator: (-1)^n, (-1)^(n-1), cos(nPi) (b_n)=the Absolute Value of (a_n) if a)lim(b_n) as n->infinity=0 b) b_n is decreasing then the Sum of A_n is convergent
Proof by induction Specific case to general a)Show decreasing between a_1 & a_2 b)assume sequence is decreasing with general terms for some a_(k+1)<a_k c) show decreasing for the (k+2) case
Integral Test if f(x) is a function where f(n)=An for n># AND a)f cont for x># b)f is T term for x>=# c)f is decr for x>=# THEN a)Int from 1->infinity of f(x)dx Conv implies Sum of a_n from n=1->inf. conv b)Int from 1->inf div impl. Sum n=1->inf a_n also div
p-series E n=2->inf 1/n^p<Int 1->inf 1/(x^p)dx<E n+1->inf 1/n^p INT 1->inf. 1/(x^p)dx converges for p>1 and diverges for p=<1
Limit Comparison Test both a_n and b_n must be (+) term Lim n->inf (a_n)/(b_n)
Absolute/conditional convergence E ABS V a_n CONV E(a_n) is abs conv DIV Test E(a_n) *conv AST, cond conv. *div E(a_n) div
ratio Test works well on x^n and n! always fails w/ only powers of n->1/n^2, 1/n^(1/2) 1)if lim n->inf AbsV (a_n+1)/a_n=L &L>1 then Sum A_n is Abs conv 2)if lim n->inf AbsV " =L &L<1 then Sum A_n is div 3)if " " =1 then test fails
Root Test use when nth power occurs ->n^n, a^n 1)if lim n->inf AbsV (a_n)^1/n= L<1 then Sum A_n is Abs Conv 2) if " " = L>1 then Sum a_n is div 3) if " " =1 then test fails
Harmonic Series Sum n=1->inf 1/n is divergent
Telescoping Series change Sum n=1->inf An to Sum i=1->n Ai solve using partial fractions, then write out 1st and last terms until things cancel. take lim n->inf of series terms
Created by: dtbrowne7