IMPROPER INTEGRALS1∫∞ dx/xp if p > 1 convergesif p ≤1 divergesDirect 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 DINFINITE SERIES1)Check if lim an= 0 ; if NO, series is Dn∞ ∞2)Geometric? Is it of the form: ∑ ar(n-1) = a/(1-r)n=1if |r| < 1, then converges to (1-r)if |r| ≥ 1, then diverges ∞3)Telescoping? Is if of the form: ∑ (an– an+1)n=1if an 0, then sum = a1-try rewriting with partial fractions∞4)Is it of the form ∑ 1/npn=1 if p > 1 C; if p ≤ 1 D5)Direct Comparison Test: ∞If an > 0 for all n AND an ≤ cnand ∑ cnis C,∞n=1then ∑ anis C n=1∞∞if an≥ dnand ∑ dnis D , then ∑ anis Dn=1n=16)Limit Comparison Test: (put what you are comparing w/ on the bottom)If lim an/bn= c > 0, then ∑an& ∑bnboth C or Dn∞ = 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 Cn∞ >1, then D=1, inconclusive9) 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>Nii) an+1 ≤ an for n>Niii) lim an = 0n∞ -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 theratio test used to find the limit-radius is what the limit goes to-conditionally convergent where x makes CCIf ∑ an is C, then series is absolutely convergentIf ∑ (-1)n+1anis C but ∑anis D then ∑ (-1)n+1anisconditionally convergent. ∞10)Power Series: power series about b: ∑Cn(x-b)n1)Ratio Test it!n=02) get radius of convergence3)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 xinterval 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!