IMPROPER INTEGRALS 1∫∞ dx/xp if p > 1 converges if p ≤1 diverges Direct Comparison: g(x) ≥ f(x) if ∫g(x)dx C, so does ∫f(x)dxif ∫f(x)dx D, so does ∫g(x)dxLimit Comparison:If lim g(x)= L then a∫∞f(x)dx and a∫∞g(x)dx x∞ f(x) both C or D INFINITE SERIES 1)Check if lim an= 0 ; if NO, series is D n∞ ∞2)Geometric? Is it of the form: ∑ ar(n-1) = a/(1-r) 3)Telescoping? Is if of the form: ∑ (an–an+1) 4)Is it of the form ∑ 1/npn=1 if p > 1 C; if p ≤ 1 D 5)Direct Comparison Test: ∞If an > 0 for all n AND an ≤ cnand ∑ cnis C, ∞n=1 then ∑ anis C n=1∞∞if an≥ dnand ∑ dnis D , then ∑ anis D n=1n=1 6)Limit Comparison Test: (put what you are comparing w/ on the bottom) If lim an/bn= c > 0, then ∑an& ∑bnboth C or D n∞ = 0 then C of ∑bnguarantees C of ∑an= ∞ then D of ∑bnguarantees D of ∑an7)Ratio Test: (good w/ n!’s)If lim an+1/an<1, then C n∞ >1, then D =1, inconclusive 9) Alternating Series Theorem(check absolute convergence first, then use i) ii) iii) to check for conditional convergence) ∞∑(-1)nan converges if: n=1 i) an > 0 for n>N ii) an+1 ≤ an for n>N iii) lim an = 0 n∞ -interval of C is when the end points are checked (if converges, then is also equalto the endpoint) -absolute interval of C is the interval given by the ratio test used to find the limit -radius is what the limit goes to -conditionally convergent where x makes CC If ∑an is C, then series is absolutely convergentIf ∑ (-1)n+1anis C but ∑anis D then ∑ (-1)n+1anis conditionally convergent. ∞10)Power Series: power series about b: ∑Cn(x-b)n 1)Ratio Test it!n=02) get radius of convergence 3)Test end points (plug x’s back into original series & see if series converges with a test): if C, then (=) in interval of convergence and converges conditionally at that x interval of absolute conv. Stays same exceptif end point x causes absolute convergence (in alt. series) ∞11)Taylor Series: for f(x) about x=a is ∑ f(n)(a)(x-a)nn=0n!