Alternating Harmonic Series:
𝑛 1 , By the Alternating
∑∞
𝑛→∞(−1)
𝑛
Absolute Convergence: If the series
∞
∑∞
𝑛→1|𝑎𝑛 |, then the series ∑𝑛→1 𝑎𝑛
converges.
1
𝑛
𝑛1
∑∞
𝑛→∞(−1) 𝑛
Series Test, lim𝑛→∞ = 0 and
1
𝑛+1
1
𝑛
< , therefore
converges.
Ratio Test: Consider lim𝑛→∞ |
𝑎𝑛+1
|
𝑎𝑛
=
Conditional Convergence: ∑∞
𝑛→1 𝑎𝑛 is
conditionally convergent if ∑∞
𝑛→1|𝑎𝑛 |
diverges, but the series converges
𝐿
*Works BEST with powers of n or
factorials
If L < 1, then the series absolutely
converges.
𝑛
Root Test: Consider lim𝑛→∞ √|𝑎𝑛 | =
𝐿
If L > 1, then the series diverges.
If L < 1, then the series absolutely
converges.
If L = 1, then try another test.
If L > 1, then the series diverges.
If L = 1, then try another test.
𝑛
Power Series: Consider ∑∞
𝑛→1 𝑎(𝑥 − 𝑏)
If |𝑥 − 𝑏| < 1, then the series converges.
Telescoping Series: By using
lim𝑛→∞ 𝑆𝑛 , we can use the partial
sums to write a sequence definition
for the partial sums.
Sequences: Use lim𝑛→∞ 𝑎𝑛 and find
the limit.
If the limit exists, then the sequence
converges.
If the limit doesn’t exist (equals
infinity or doesn’t approach
anything), then the sequence
diverges.
Sequences
Monotonic: Either always increasing
or always decreasing
Bounded below: lower limit
Bounded upper: upper limit
𝑛
Geometric Series: ∑∞
𝑛=1 𝑎𝑟
Divergence Test
lim𝑛→∞ 𝑎𝑛 = 0
|r| < 1
Move on to different
test
Converges
lim𝑛→∞ 𝑎𝑛 ≠0
Diverges
|r| ≥ 1
Diverges
𝑎
1
Sn= 1−𝑟
P-Series: ∑∞
𝑛=1
P>1
Integral Test: If ∑∞
𝑛=1 𝑎𝑛 whose terms
are positive and decreasing, then if
∞
∫1 𝑓(𝑛)dn converges, so does ∑∞
𝑛=1 𝑎𝑛
1
𝑛𝑝
P≤1
Comparison Test: Generally, compare
to P-Series or Geometric Series.
Converges
Diverges
If ∑∞
𝑛=1 𝑎𝑛 is less than a convergent
∞
series ∑∞
𝑛=1 𝑏𝑛 , then ∑𝑛=1 𝑎𝑛 also
converges.
If ∑∞
𝑛=1 𝑎𝑛 is greater than a divergent
∞
series ∑∞
𝑛=1 𝑏𝑛 , then ∑𝑛=1 𝑎𝑛 also
diverges.
Limit Comparison Test: Explore the
limit of the ratios of “our series” and
some known series.
𝑎𝑛
= 𝑐, 𝑐 > 0, then the
𝑏𝑛
∞
∑𝑛=1 𝑎𝑛 and ∑∞
𝑛=1 𝑏𝑛 , either
lim𝑛→∞
series
both converge or diverge.
Alternating Series: The Alternating
𝑛
Series ∑∞
𝑛→∞(−1) 𝑎𝑛 and
𝑛+1 𝑎 converge if the
∑∞
𝑛→∞(−1)
𝑛
following conditions are met.
1. lim𝑛→∞ 𝑎𝑛 = 0
2. 𝑎𝑛+1 < 𝑎𝑛