calc c - series and sequences

Harmonic series

1/n diverges

Geometric series

ar^n a series where all sequencial terms yeild a common ratio. Sn = a(1-r^n)/(1-r) so converges to a/(1-r) when -1<r<1

subtraction trick

write the nth partial sum generally (Sn = a1 + a2 + a3 + … + an) and then create another identical term but multiply both sides by some value and subtract. this will usually leave the first and end terms as a simple answer to nth partial sum

sequence of terms

a list of all the terms that will be added together in the series.

sequence of partial sums

a list of all the partial sums up to some n value (S5 is the sum of all the first 5 term where n=5). the limit of this sequence can tell us if the series converges or diverges

sequence convergence rules

1/n^p converges when p>0, r^n = {0 (-1<r<1), 1 (r=1), diverges otherwise}

sequence vs function rule

for a sequence that follows a continuous curve described by f(x) if the limit of f(x) = L then the limit of the sequence is L

p-series

form 1/n^p where p is positive. converges p>1, diverges p<=1

Integral Test

If 1) F(x) continuous on [1,infinity), 2) f(x) decreasing on [1,infinity), 3) f(n) = an for n>=1

then series converges/diverges when integral converges/diverges

test for divergence

if lim(n→infinity) an does not equal 0, then the series diverges. if it equals 0 this test can’t tell us if it converges (it might diverge in that case too)

regular comparison test

test a series we don’t know with a series we do know converges/diverges. depending on the size of the series it will force the unknown to converge or diverge. prove the relation of size

limit comparison test

supposing An and Bn are positive series for all n, if lim(n→infinity) An/Bn = C where c is positive and finite. basically pick Bn that acts like An that you know converges or diverges

telescoping series

most of the terms in the nth sum cancel out making limit easy to find. often when there is a term subtracted from another

alternating series test

if 1) decreasing and positive 2) lim bn = 0, then alternating series converge

convergence types

a series like an alternating series can converge or diverge (by alternating series test) and its absolute value may do something differnet (by another test). impossible for series to diverge while absolute value converges

ratio test

p = lim |An+1/An|, if 0<=p<1 converges absolutely, if p>1 diverges, if p=1 test inconclusive

root test

p = lim |An|^(1/n), if 0<=p<1 converges absolutely, if p>1 diverges, if p=1 test inconclusive. behaves like geometric sequence with r=roe

Term-by-term differentiation/integration

if you can turn 1/(1-x) into some function f(x) through substitution, multiplication, diff/int. you can take the series x^n, use the same steps, and find the coresponding series for f(x)

how to find convergence radius/interval of a series

use ratio test, if 0<p<infinity, sub result in for x in -1<x<1 and simplify. test the end points. radius is distance from end point to center

1/(1-x) coresponding series

x^n

sinx power series

x^1/1! - x³/3! + x^5/5! - x^7/7! + …

cosx power series

x^0/0! - x^2/2! + x^4/4! - x^6/6! + …

e^x power series

x^0/0! + x^1/1! + x^2/2! + x^3/3! + …