Series Tests

p-series

∑ {n=1,∞} 1/n^p

convergent if p>1

divergent if p<=1

geometric series

a + a*r + a*r + a*r² + a*r³ ... converges to [a/1-r]

convergent if r<1

divergent if r>=1

Strategy: write out the first n terms leaving exponents in place, find what r is, in other words find what is being multiplied by the previous term.

comparison test

if ∑an and ∑bn are series with positive terms

........if ∑an <= ∑bn and ∑bn converges, ∑an converges

........if ∑an <= ∑bn and ∑an diverges, ∑bn diverges

if the sum of a series is less than another series but the functions are about the same, then if the smaller one diverges the one above must also diverge, if the bigger one converges, then the smaller one must converge also

∑(n⁵+3n²+42)/(n⁶√(n)+1) ~ ∑n⁵/n⁶ ~ ∑1/n³/² which is a p-series

limit comparison test

if lim {n→∞} an/bn = c , where 0

......then ∑an converges if and only if ∑bn converges

......then ∑an diverges if and only if ∑bn diverges

if a is sometimes positive and sometimes negative we can use ∑|an|

test for divergence

if lim {n→∞} an =/= 0

......then ∑an diverges

alternating series

b₁-b₂+b₃-b₄+b₅-b₆ where bn >= 0

∑(-1)ⁿ bn

.......if bn+1 <= bn (if decreasing)

.......if lim {n→∞} bn = 0

then (-1)ⁿ bn converges

ratio test

lim |(an+1)/(an)| {n→∞}

(for n+1, just subsitute in n+1 for every n)

if L<1 absolutely convergent and convergent

if L>1 divergent

if L=1 you know nothing

root test

if lim {n→∞} ⁿ√|an|

if lim < 1, an conveges

if lim >1 an diverges

if lim = 1 you know nothing

integral test

if an = f(n)

if ∫ {1,∞} f(x) converges then ∑an converges

if ∫{1,∞} fx diverges then ∑an diverges

absolutely convergent

the sum of the absolute value of an is finite

conditionally convergent

if a sum is convergent but is not absolutely convergent