Series Rules

The divergence test

lim⁡n→∞an\lim_{n\to\infty} a_n

  • If lim⁡n→∞an≠0⇒∑an\lim_{n\to\infty} a_n \ne 0 \Rightarrow \sum a_n Diverges

  • If lim⁡n→∞an=0⇒∑an\lim_{n\to\infty} a_n = 0 \Rightarrow \sum a_n may converge or diverge and you need to do another test

Geometric Series

∑n=1∞a(r)n−1\sum_{n=1}^\infty a(r)^{n-1}

  • If |r| < 1 , The series converges

  • If ∣r∣≥1|r| \ge 1, The series diverges

P-Series Test

In the form of ∑n=1∞1np\sum_{n=1}^\infty \frac 1{n^p}

  • If p > 1, the series converges

  • If p≤1p \le 1, the series diverges

Telescoping Series

ex)

∑n−1∞=1−12+12−13+13−14+…\sum_{n-1}^\infty = 1 - \frac 12 + \frac 12 - \frac 13 + \frac 13 - \frac 14 + …

You can cancel out the terms in the middle and this will go to the general form ana_n and you want to write a formula that will give you the partial sum of the telescoping series.

∑n−1∞an=lim⁡n→∞Sn\sum_{n-1}^\infty a_n = \lim_{n\to\infty} S_n

  • If you get a finite value, the series converges.

  • If you get something like ±∞\pm\infty or DNE, the series diverges

ex)

∑n−1∞(1n−1n+1)\sum_{n-1}^\infty (\frac 1n - \frac 1{n+1})

⇒(11−12)+(12−13)+(14−15)+…\Rightarrow (\frac 11 - \frac 12) + (\frac 12 - \frac 13) + (\frac 14 - \frac 15) + …

Integral Test

an=f(n)a_{n}=f(n)

  • ff has to be positive, continuous, and decreasing function from [n,∞)[n,\infty)

∫1∞f(x)dx=L\int_1^\infty f(x)dx = L where L is some finite value

  • If LL, the series is convergent

  • If ±∞\pm \infty or DNE, divergent

Ratio Test

\lim_{n\to\infty} | \frac{a_{n+1}}{a_n}| < 1 \Rightarrow Converges

\lim_{n\to\infty} | \frac{a_{n+1}}{a_n}| > 1 or +∞⇒+\infty \Rightarrow Diverges

lim⁡n→∞∣an+1an∣=1⇒\lim_{n\to\infty} | \frac{a_{n+1}}{a_n}| = 1 \Rightarrow Inconclusive

Root Test

\lim_{n\to\infty}\sqrt[n]{|a_{n}|} < 1 \Rightarrow Converges

\lim_{n\to\infty}\sqrt[n]{|a_{n}|} > 1 \Rightarrow Diverges

lim⁡n→∞∣an∣n=1⇒\lim_{n\to\infty}\sqrt[n]{|a_{n}|} = 1 \Rightarrow Inconclusive

Direct Comparison Test

0≤an≤bn0 \le a_n \le b_n

  • If ∑bn→\sum b_n \rightarrow Converges, ∑an→\sum a_n \rightarrow Converges

  • If ∑an→\sum a_n \rightarrow Diverges, ∑bn→\sum b_n \rightarrow Diverges

Limit Comparison Test

If lim⁡n→∞anbn=L\lim_{n\to\infty} \frac {a_n}{b_n} = L both will converge or diverge

  • If ∑an→\sum a_n \rightarrow Converges, ∑bn\sum b_n will also converge due to L being a finite number

  • If ∑an→\sum a_n \rightarrow Diverges, ∑bn\sum b_n will also Diverge

Alternating Series Test

∑n=1∞(−1)nan→\sum_{n=1}^\infty (-1)^n a_n \rightarrow Converges if:

  • lim⁡n→∞an=0\lim_{n\to\infty} a_n = 0

  • an≥an+1≥an+2+…a_n \ge a_{n+1} \ge a_{n+2} + …

Absolute Value Test

If ∑∣an∣→\sum |a_n| \rightarrow Converges

  • Then ∑an→\sum a_n \rightarrow Converges

Which means it is absolutely convergent

If ∑∣an∣→\sum |a_n| \rightarrow Divergent and ∑an→\sum a_n \rightarrow Convergent, then the original series is conditionally convergent