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44 Cards in this Set
- Front
- Back
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Is the SEQUENCE convergent or divergent? -- STEPS |
> Set "an=" to "lim(n to inf)" > divide by largest n in denominator > limit exists = convergent. Otherwise, divergent |
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For what values of R is the sequence convergent? -- STEPS |
> find value of R that eliminates infinities > usually between -1 and 1 (decimal) |
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Is the SERIES convergent or divergent? -- OPTIONS |
> check for Geometric Series (ar^n-1) > check for partial sums > test for divergence (lim != 0) > integral test ( lim( integral( f(x) ) ) ) > check for p-series (1/n^p) > comparison test (an <= geo-/p-series) > limit comparison test (lim(an / bn)) |
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Find the sum of the geometric series [series]. -- STEPS |
> SUM(n=1 to inf) ar^(n-1) converts to: a / (1-r) >> a = the value at n=1 >> r = the common ratio >>> replace a and r, then solve |
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Find the sum of the series [series]. -- STEPS |
> check for Geometric Series > check for partial sums > manually solve |
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The Integral Test |
> f is a continuous, positive, decreasing function on [1, inf) > let: an = f(n) >> if integral(1 to inf) f(x) dx exists, then SUM(n=1 to inf) (an) is convergent >> if nonexistant, SUM(n=1 to inf) (an) is divergent
Can also switch integral(1 to inf) with lim(t to inf) integral(1 to t) |
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Test for Divergence |
> if lim(n to inf) (an) doesn't exist or if lim(n to inf) (an) != 0, the series is divergent. |
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Partial Sums Form |
> split an into 2 parts
EXAMPLE > SUM(n=1 to inf) (an) = 1 / (n*(n+1) > lim(n to inf) 1 - 1/(n+1) >> = 1-0 = 1 |
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P-Series |
> SUM(n=1 to inf) (1/n^p) >> convergent if p > 1 >> divergent if p <= 1 |
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Remainder Estimate for Integral Test |
> f(k) = ak; f is a continuous, positive, decreasing function for x >= n; SUM (an) is convergent Integral(n+1 to inf) f(x) dx <= Rn /*s-sn*/ <= Integral(n to inf) f(x) dx |
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a) Approximate the sum of the series [series] by using the sum of the first 10 terms. Estimate the error. b) How many terms are required to ensure that the sum is accurate to within [margin] -- STEPS |
> ITR = Integral(n to inf) f(x) dx a) add first 10 in series > plug 10 into ITR = [margin] b) solve inequality ITR < margin > round answer to get even n |
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Series Total Sum Using First "n" Terms, Estimation Formula |
Sn + Integral(n+1 to inf) f(x) dx <= S <= Sn + Integral(n to inf) f(x) dx |
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Use n = 10 to estimate the sum of the series [series]. -- STEPS |
> check for geometric comparison, otherwise: > find Sn > solve Integral(n to inf) f(x) dx > Use "Total Sum from the First n Terms Estimation" > find midpoint of two sides > margin of error is less than ITR |
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The Comparison Test |
If SUM(an) is being compared to SUM(bn): > if bn is convergent, an must be smaller than bn for all n >> an is convergent if this is true > if bn is divergent, an must be greater than bn for all n >> an is divergent if this is true |
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Limit Comparison Test |
> lim(n to inf) an/bn = C > if C > 0: >> bn is p-series = an converges if (bn)p > 1 >> bn is geo-series = an converges if (bn)|r| < 1 >> bn is harmonic = an converges |
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Use the Integral Test to determine whether the series [series] is convergent or divergent. -- STEPS |
> f(an) = f(x) dx > integral(n to inf) f(x) dx > use lim(t to inf) integral(n to t) f(x) dx > solve; if limit is convergent, so is SUM >> if limit is divergent, so is SUM |
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Estimating Sums by Comparison |
> compare [an] to similar series [bn] > integral test to get [bn]'s error at [n] > find an's approximation at n |
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Use the Limit Comparison Test to determine if the series [series] converges or diverges. -- STEPS |
> find dominant parts in numerator / denominator > set lim(n to inf) an/bn = C > simplify; divide by highest n in denominator > if C > 0 and finite, (an) con/diverges (whatever it is for bn) |
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Alternating Series Test |
If the alternating series bn*(-1)^n-1 satisfies > b(n+1) <= bn, for all n > lim(n to inf) bn = 0 Then the series is convergent |
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Test the alternating series [series] for convergence or divergence. -- STEPS |
> verify that b(n+1) < bn >> if not obvious, use L'Hopital's Rule >>> find x that makes f'(x) <= 0 >>> n must be int after that number >>otherwise, Test for Divergence > verify lim(n to inf) bn = 0 |
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Alternating Series Estimation Theorem |
If S = SUM(bn*(-1)^(n-1)) is the sum of the alternating series that satisfies > b(n+1) <= b > lim(n to inf) bn = 0 Then |Rn| = |S-Sn| <= b(n+1) |
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Find the sum of the alternating series [series] correct to [x] decimal places. -- STEPS |
> check each n until bn passes over the goal (e.g. 1/6! > .001, but 1/7! < .0002 when goal was 3 decimal places) > calculate S(n-1) > S - S(n-1) <= bn < target (rounded bn) |
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Absolutely Convergent vs Conditionally Convergent |
A series is: > absolutely convergent if SUM|an| is convergent (|an| at all n) > conditionally convergent if SUM(an) is convergent, but SUM|an| is not |
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The Ratio Test |
(i) If lim(n to inf)|a(n+1) / an| < 1, (an) is absolutely convergent (ii) if --the above limit-- > 1 or = inf, (an) is divergent (iii) if --the above limit-- = 1, no conclusion can be drawn about the con/divergence of (an) |
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Use the Ratio Test to test the series [series] for convergence or divergence. -- STEPS |
> make a fraction with a(n+1) plugged in on top and an on the bottom > evaluate lim(n to inf) > if lim >= 1, test for divergence |
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The Root Test |
(i) if lim(n to inf)(nth_root(|an|)) < 1, the series is absolutely convergent (ii) if --the above limit-- > 1 or = inf, the series is divergent (iii) if --the above limit-- = 1, Root Test is inconclusive |
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Test the convergence of the series [series] using the Root Test. -- STEPS |
> remove root if existing, and take absolute value of an > evaluate limit |
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SUM(n to inf) 1/n^p |
P-series: P > 1 = convergent P <= 1 = divergent |
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SUM(n to inf)(ar^(n-1)) |
Geometric series: |r| < 1 = convergence |r| > 1 = divergence *may need preliminary conversions |
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SUM(n to inf)(an) similar to p-series or geometric series. |
P-Series: > Rational / algebraic function > keep only highest powers of n in numerator & denominator Comparison tests only apply to positive terms; if not positive, test for absolute convergence. |
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SUM(n to inf)(an) looks like lim != 0 |
Test for Divergence: 0 = convergence !0 = divergence |
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SUM(n to inf)(-1)^(n-1) or ^n |
Alternating Series Test: > bn >= b(n+1) for all n > lim(n to inf) bn = 0 If both are true, convergence Else, divergence |
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SUM(n to inf)(an) -- factorials, constants raised to nth power |
Ratio Test: > NOT used on p-series (rational / algebraic functions of n) > lim < 1 = convergence > lim > 1 or inf = divergence > lim = 1 = inconclusive |
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SUM(n to inf)(bn)^n |
Root Test: > lim < 1 = convergence > lim > 1 or inf = divergence > lim = 1 = inconclusive |
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SUM(n to inf)(an) easily evaluated as integral(1 to inf)(f(x) dx) |
Integral Test > positive, continuous, decreasing on [1, inf) > f(x) convergent = an convergent > f(x) divergent = an divergent |
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For what values of x is the series SUM(n to inf) (n! * x^n) convergent? -- STEPS |
> has factorial, use Ratio Test >> an = n! * x^n >> lim(n to inf)|a(n+1) / an| = inf (when x != 0) > converges only at x = 0 |
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For what values of x is the series SUM(n to inf) (x-3)^n / n convergent? -- STEPS |
> Ratio Test >> an = (x-3)^n / n >> lim|a(n+1) / an| = |x-3| >>> convergent when |x-3| < 1 >>> 2 <= x < 4 |
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For a given Power Series about A [series], there are only 3 possibilities. |
(i) The series converges only when x=a (ii) The series converges for all x (iii) There is a positive number R (Radius of Convergence) such that the series converges if |x-a| < R and diverges if |x-a| > R |
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Interval of Convergence |
Interval that consists of all values of x for which the series converges Case (i): single point, a Case (ii): interval (-inf, inf) Case (iii): 4 possibilities 1. (a-R, a+R) 2. (a-R, a+R] 3. [a-R, a+R) 4. [a-R, a+R] |
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Determine the radius of convergence and the interval of convergence in the series [series]. -- STEPS |
> use Ratio or Root test > find limit in terms of x > whatever |x-a| must be to satisfy limit = 1 is the Radius of Convergence, R > interval of convergence for all values of x in range |x-a| < 1 |
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Express function [function] as the sum of a power series and find the IoC. -- STEPS |
> convert to different form (geometric? Look for identities) > need some form of Cn*x^n > to find IoC, look for clues (e.g. geometric form -- |x| < 1 > find x and express as an interval of convergance |
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Term-By-Term Differentiation and Integration |
If the power series [PS about A] has a RoC > 0, then the function defined by C0 + C1(x-a) + C2(x-a)^2 ... Sum(Cn*(x-a)^n) is differentiable (and continuous) on the interval (a-R, a+R), and: (i) f'(x) = C1 + 2C2(x-a) + 3C3(x-a)^2 ... Sum(n*Cn*(x-a)^(n-1)). (ii) integral(f(x) dx) = Ct + C0(x-a) + C1((x-a)^2 / 2) + C3((x-a)^3 / 3) ... Ct + Sum(Cn * ((x-a)^(n+1) / (n+1))) *RoC stays the same |
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Find a power series representation for [function] and its Radius of Conversion. -- STEPS |
> write out the series of the function consistent with a known type (geometric?) > differentiate the function and write it as an integral > the entire series gets written as an integrand one by one (do 5 then ...) dx > integrate one by one with C for constant > express the new series in summation notation > to get C, fill x=0 into original function |
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Power Series Forms |
Forms: > SUM(n to inf) Cn*x^n > SUM(n to inf) Cn*(x-a)^n >>power series centered at / about a |
