If limn→∞an=0⇒∑an may converge or diverge and you need to do another test
Geometric Series
∑n=1∞a(r)n−1
If ∣r∣<1 , The series converges
If ∣r∣≥1, The series diverges
P-Series Test
In the form of ∑n=1∞np1
If p>1, the series converges
If p≤1, the series diverges
Telescoping Series
ex)
∑n−1∞=1−21+21−31+31−41+…
You can cancel out the terms in the middle and this will go to the general form an and you want to write a formula that will give you the partial sum of the telescoping series.
∑n−1∞an=limn→∞Sn
If you get a finite value, the series converges.
If you get something like ±∞ or DNE, the series diverges
ex)
∑n−1∞(n1−n+11)
⇒(11−21)+(21−31)+(41−51)+…
Integral Test
an=f(n)
f has to be positive, continuous, and decreasing function from [n,∞)
∫1∞f(x)dx=L where L is some finite value
If L, the series is convergent
If ±∞ or DNE, divergent
Ratio Test
limn→∞∣anan+1∣<1⇒ Converges
limn→∞∣anan+1∣>1 or +∞⇒ Diverges
limn→∞∣anan+1∣=1⇒ Inconclusive
Root Test
limn→∞n∣an∣<1⇒ Converges
limn→∞n∣an∣>1⇒ Diverges
limn→∞n∣an∣=1⇒ Inconclusive
Direct Comparison Test
0≤an≤bn
If ∑bn→ Converges, ∑an→ Converges
If ∑an→ Diverges, ∑bn→ Diverges
Limit Comparison Test
If limn→∞bnan=L both will converge or diverge
If ∑an→ Converges, ∑bn will also converge due to L being a finite number
If ∑an→ Diverges, ∑bn will also Diverge
Alternating Series Test
∑n=1∞(−1)nan→ Converges if:
limn→∞an=0
an≥an+1≥an+2+…
Absolute Value Test
If ∑∣an∣→ Converges
Then ∑an→ Converges
Which means it is absolutely convergent
If ∑∣an∣→ Divergent and ∑an→ Convergent, then the original series is conditionally convergent