IMPROPER INTEGRALS1∫∞dx/xpif p > 1convergesif p ≤1divergesDirect 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 thena∫∞f(x)dx anda∫∞g(x)dxx∞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 an0, then sum = a1-try rewriting with partial fractions∞4)Is it of the form ∑ 1/npn=1if p > 1C; if p ≤ 1D5)Direct Comparison Test:∞If an> 0 for all n AND an≤ cnand ∑ cnis C,∞n=1then ∑ anis Cn=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)nanconverges if:n=1i) an> 0 for n>Nii) an+1≤ anfor 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 ∑ anis 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 ifseries converges with a test):if C, then (=) in interval of convergence and convergesconditionally at that xinterval of absolute conv. Stays same exceptif end point xcauses absolute convergence (in alt. series)∞11)Taylor Series: for f(x) about x=a is ∑ f(n)(a)(x-a)nn=0n!
