ch 12 test WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

explicit and recursive equations for arithmetic and geometric sequences and series

• arithmetic

  • sequence

    • explicit: tn = t1 + (n-1) d

    • recursive: tn = tn-1 + d

  • series

    • explicit: n(t1 +t2)/2

    • recursive: u(n) = u(n-1)+(u(nMin-1) + dn)

• geometric

  • sequence

    • explicit: tn = t1 * r ^ n-1

    • recursive: tn = (tn-1)r

  • series

    • explicit: t1(1-r^n)/1-r

    • recursive: u(n) = u(n) = u(n-1) + u(nMin)(r ) ^n

when does a geometric infinite series converge

when the abs of r is less than 1

what value does it converge to

t1 over 1-r

partial sum of geometric series

t1 times 1-r^n over 1-r

power series formula

f^n(a)

sigma —--—- (x-a)^n

n!

eight basic power series

what does root technique prove

absolute convergence and therefore regular convergence

list of tests

1. nth term test

2. harmonic series

3. p series test

4. geometric series test

5. alternating series test

6. integral test

7. ratio test

8. direct comparison test

9. limit comparison test

conditions of nth term test and what does it prove

conditions: none

proves: divergence

if the limit does not equal 0, it diverges

conditions of harmonic test and what does it prove

conditions: terms pos

proves: divergence

pos reciprocal of arithmetic series means it diverges

conditions of p series test and what does it prove

conditions: terms are pos

proves: convergence and divergence

if p > 1, converges, if p is <= 1, diverges

conditions of geometric test and what does it prove

conditions: summand is geometric

proves: convergence and divergence and convergence value

if |r| < 1, converges to t1 / 1-r

else, diverges

conditions of alternating test and what does it prove

conditions: terms alternate in sign, |tn| decreases, lim as n goes to infinity is 0

proves: convergence

if conditions are met, it converges. else, indeterminate

conditions of integral test and what does it prove

conditions: pos, terms decrease, and continuous

proves: convergence and divergence

whichever the integral converges/diverges, same happens to og function

conditions of ratio test and what does it prove

conditions: none

proves: convergence and divergence

if L is the limit as n approaches infinity of |tn+1 / tn| and L < 1 converges absolutely, if L > 1 diverges, else indeterminate

conditions of direct comparison test and what does it prove

conditions: an + bn > 0

proves: convergence and divergence

if bound above by convergence then converges, if bound below by divergence then diverges

else indeterminate

conditions of limit comparison test and what does it prove

conditions: f(n) > 0 and g(n) > 0

proves: convergence and divergence

if L > 0 then f(n) converges however g(n) does

if L = 0 and g(n) converges, then f(n) converges

if L = infinity and g(n) diverges, then f(n) diverges

convergence with combined series

• if two combined series separately converge, then sum converges

• if one series converges but the other diverges, then sum diverges

• if both diverge, then indeterminate

estimating |Rn| with alternating series

if an infinite series converges by alternating series test, then |Rn| <= |tn+1| (the next term in the series)

estimating with integral test

|Rn| will be <= the | integral from the index of the partial sum (same index) to infinity |

• first have to show series passes conditions of integral test

error equations

1. |Rn| = |f(b) - Pn(b)|

2. |Rn| <= |f^n+1(c ) |

   | ——— (b-a) ^ n+1 |

   | (n+1)! |

how to find series of PS from seq of PS

• the difference between succeeding terms in a seq of PS is the values of the par series

• so just take the differences between the sequence terms and those are the terms of your partial series

don’t forget

• PLUS C

• show partial sum for telescoping series