Flashcards
Calculus - Chapter Two
Cards in this deck (20)
- 1SERIES Suppose that a1,a2, ... elements of R. The series SIGMA(n=1,infinity)[an] is ... If sn = a1+a2+a3+...+an = SIMGA(i=1,n)[ai] what is sn called? What is an in this? If {sn} converges with limit s then ... If {sn} does not converge then ...... the infinite sum a1+a2+a3+...
sn is called the nth partial sum, where an is the nth term
... SIGMA(n=1,infinity)[an] converges with sum s
... SIMGA(n=1,infinity)[an] diverges - 2Suppose that SIGMA(n=1,infinity)[an] and SIGMA(n=1,infinity)[bn] converge with sums A and B, and suppose alpha,beta elements of R. Then a) SIGMA(n=1,infinity)[alpha*an] converges with sum ... b) SIGMA(n=1,infinity)[alpha*an + beta*bn] converges with sum ...... alpha*A
... alpha*A + beta*B
PROOF ON SEPARATE SHEET!! - 3Adding, removing or changing a finite number of terms in a series does not affect what? What happens when series are subtracted from one another?The convergence or divergence
The limits are subtracted from one another - 4If SIGMA(n=1,infinity)[an] converges then ... PROOF ON SEPARATE SHEET!!... an tends to 0 as n tends to infinity
- 5If an tends to 0 as n tends to infinity then ... Does SIGMA(n=1,infinity)[1/n] converge or diverge ? If an does not tend to 0 as n tends to infinity then ... PROOF ON SEPARATE SHEET!!this does not mean that SIGMA(n=1,infinity)[an] converges
Diverge
SIGMA(n=1,infinity)[an] diverges - 6SERIES WITH NON-NEGATIVE TERMS If an>/0 for each n, then the sequence of partial sums {sn} is ... and SIGMA(n=1,infinity)[an] ... if and only if ... PROOF ON SEPARATE SHEET!!... (monotonically) increasing ... converges ... {sn} is bounded above
- 7When {sn} is bounded above the sum of SIGMA(n=1,infinity)[an} is ...... the lub (lower upper bound) of {sn}
- 8State the comparison test PROOF ON SEPARATE SHEET!!Suppose 0
- 9Suppose alpha element of R with alpha >0. State the conditions where SIMGA(n=1,infinity)[1/n^alpha] a) converges b) divergesa) converges if alpha >1
b) diverges if alpha - 10State the Alternating Series Test PROOF ON SEPARATE SHEETSuppose that {un} is a decreasing sequence with un>/0 and that un tends to 0 as n tends to infinity.
Then SIGMA(n=1,infinity)[(-1)^(n-1)*un converges - 11State the Ratio Test for ConvergenceSuppose that an>0 for every n and suppose that (an+1/an) tends to l as n tends to infinity
a) If l<1 then SIGMA(n=1,infinity)[an] converges
b) If l>1 then SIGMA(n=1,infinity)[an] diverges - 12ABSOLUTELY CONVERGENT SERIES The series SIGMA(n=1,infinity)[an] is said to be absolutely convergent if ... If SIGMA(n=1,infinity)[an] is absolutely convergent then ... PROOF ON SEPARATE SHEET!!... SIGMA(n=1,infinity)[modulus(an)] is convergent
... it is convergent - 13State the Ratio Test for Absolute Convergence PROOF ON SEPARATE SHEET!!Suppose that an doesn't =0 for every n and suppose that modulus(an+1)/modulus(an) tends to l as n tends to infinity.
a) If l<1 then SIGMA(n=1,infinity)[an] is absolutely convergent (so convergent)
b) If l>1 then SIGMA(n=1,infinity)[an] diverges - 14It is a property of convergent sequences that if (an+1)/an tends to l as n tends to infinity then ... A series is conditionally convergent if ...... modulus(an+1)/modulus(an) = modulus(an+1/an) tends to modulus(l) as n tend to infinity
... it is convergent but not absolutely convergent - 15POWER SERIES What is a power series? A power series usually starts with the term ...A power series is a series of the form SIGMA(n=0,infinity)[an*x^n], where {an}(n=0,infinity) is a sequence of constants and x is a real variable
... a0 - 16The MacLaurin series (e^x) is an example of the power series ... The power series SIGMA(n=0,infinity)[an*x^n] converges when ... PROOF ON SEPARATE SHEET!!... 1+x+x^2/2!+..., i.e. SIGMA(n=0,infinity)[an*x^n] where an = 1/n!
... x=0 (whatever the value of an) - 17The number R is called what? When the series converges for x=0 we write ... When the series converges for all values of x we write ...R = radius of convergence
... R=0
... R=infinity - 18Given a power series SIGMA(n=0,infinity)[an*x^n], suppose that an doesn't =0 for every n and suppose that modulus(an+1)/modulus(an) tends to a real number or infinity as n tends to infinityQuestions on next
- 19a) If modulus(an+1)/modulus(an) tends to 0 as n tends to infinity then the radius of convergence is ... b) a) If modulus(an+1)/modulus(an) tends to infinity as n tends to infinity then the radius of convergence is ...a) infinity
b) 0 - 20c) a) If modulus(an+1)/modulus(an) tends to l (which doesn't =0) as n tends to infinity then the radius of convergence is ... PROOF ON SEPARATE SHEET!!c) 1/l
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- Flashcards on Calculus - Chapter Two, created by Cara Wales on 15/05/2016.
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- Cara Wales
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