Series

Let (un)(u_n) be a sequence, and a series (a new sequence) can be defined from it such that:

sn=∑k=1nuks_n=\sum_{k=1}^{n}{u_k}

Convergence

If (sn)(s_n) is converging:

lim⁡n→∞sn=lim⁡n→∞∑k=1nuk=∑k=1∞uk=S∈R\lim_{n\to \infty}{s_n} = \lim_{n\to \infty}{\sum_{k=1}^n {u_k}} = \sum_{k=1}^\infty u_k = S\in\mathbb{R}

Absolutely Converging

∑k=1nuk\sum_{k=1}^{n}{u_k} is absolutely converging iff ∑k=1n∣uk∣ is converging\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is converging}.

∑k=1n∣uk∣ is converging  ⟹  ∑k=1nuk is converging\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is converging} \implies \sum_{k=1}^{n}{u_k}\text{ is converging}

A series sns_n is absolutely converging to ss iff rearranged series of sns_n converges to ss.

Conditionally Converging

∑k=1nuk\sum_{k=1}^{n}{u_k} is condtionally converging iff:

∑k=1n∣uk∣ is diverging      and  ∑k=1nuk is converging\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is diverging}\;\;\;\text{and}\;\sum_{k=1}^{n}{u_k}\text{ is converging}

Suppose sns_n is a conditionally converging series. Then:

  1. Sum of all the positive terms limits to ∞\infty
  2. Sum of all the negative terms limits to −∞-\infty
  3. sns_n can be rearranged to have the sum:
    • Any real number xx
    • ∞\infty
    • −∞-\infty
    • Does not exist

Terms Limit to 0

∑k=1nuk is converging  ⟹  lim⁡k→∞uk=0\sum_{k=1}^{n}{u_k}\text{ is converging} \implies \lim_{k\to\infty}{u_k} = 0

The converse is known as the divergence test:

Grouping

Suppose ∑uk\sum u_k is a given series. If vnv_n is formed by grouping a finite number of adjacent terms uku_k, then ∑vk\sum v_k is a grouping of the given series.

Rearrangement

Suppose ∑uk\sum u_k is a given series. If there is a bijection sequence knk_n such that vn=uknv_n = u_{k_n}, then ∑vn\sum v_n is a rearrangement of the given series.

Written by October 17, 2026 1 min read
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