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Series

Let (an)n=1∞(a_n)^{\infty}_{n=1} be a sequence of numbers, we define a infinite series with terms ana_n is the formal expression ∑n=1∞an\sum_{n= 1}^{\infty} a_n. Define a sequence of partial sums (sn)n=1∞(s_n)_{n=1}^{\infty} by sn=∑k=1naks_n = \sum_{k=1}^n a_k

  • The sequence of partial sums converges   ⟹  \implies this series converges or summable. Where shown in let L=lim⁡n→∞sn=∑n=1∞anL = \lim\limits_{n\to\infty} s_n = \sum_{n= 1}^{\infty} a_n
  • ∑n=1∞1n\sum_{n= 1}^{\infty} \frac{1}{n} aka the harmonic series is diverge where 1+12+⋯≥1+12+2(14)+⋯=1+n/21 + \frac{1}{2} +\cdots \ge 1 + \frac{1}{2} + 2(\frac{1}{4}) + \cdots = 1 + n/2 where use the partial sum and group from 2n−12^{n-1} to 2n2^n where always exists 2n−12^{n-1} numbers in between
  • Sn=∑n=1∞12nS_n = \sum_{n= 1}^{\infty} \frac{1}{2^n}
    • prove this converge:
    • 2Sn=1+Sn  ⟹  Sn=12S_n = 1 + S_n \implies S_n = 1

(THEOREM 3.1.4) ∑n=1∞an\sum_{n= 1}^{\infty} a_n converge   ⟹  lim⁡n→∞an=0\implies \lim\limits_{n\to\infty} a_n = 0, but verse not.

  • Which means, if we check and get the limit is not 0 we can immediately to judge it is diverge.

(CAUCHY CRITERION FOR SERIES): Let ∑i=1∞ai\sum_{i = 1}^{\infty} a_i be such series:

∑i=1∞ai\sum_{i = 1}^{\infty} a_i converges   ⟺  ∀ϵ>0,∃N∈N,s.t.∀n≥N,∣∑k=n+1∞ak∣<ϵ  ⟺  ∀ϵ>0,∃N∈N,s.t.∀n,m≥N,∣∑k=n+1mak∣<ϵ\iff \forall \epsilon > 0, \exists N\in \N, s.t. \forall n\ge N, |\sum_{k = n+1}^{\infty} a_k| < \epsilon \iff \forall \epsilon > 0, \exists N\in \N, s.t. \forall n,m\ge N, |\sum_{k = n+1}^{m} a_k| < \epsilon

(3.2.1 PROPOSITION): Consider a sequence (an)(a_n), ∀k>1,ak≥0∧sn=∑k=1nak\forall k > 1, a_k \ge 0 \land s_n = \sum_{k=1}^n a_k

  • (sn)n=1∞(s_n)_{n=1}^{\infty} is bounded above   ⟺  ∑n=1∞an\iff \sum_{n= 1}^{\infty} a_n converge
  • (sn)n=1∞(s_n)_{n=1}^{\infty} is unbound   ⟺  ∑n=1∞an\iff \sum_{n= 1}^{\infty} a_n diverge

A sequence (an)n=1∞(a_n)^{\infty}_{n=1} is a geometric sequence with ratio rr if ∀n>0,an+1=ran\forall n>0, a_{n+1} = ra_n, that is, ∀n≥0,an=a0rn\forall n \ge 0, a_n = a_0 r^n

  • geometric sequence converge when ∣r∣<1|r| < 1. And the sum of such geometric sequence is ∑n=0∞arn=a1−r\sum_{n=0}^{\infty}ar^n = \frac{a}{1-r}

Convergence Test​

(THE COMPARISON TEST): Let (an)(a_n) and (bn)(b_n) be 2 sequence of real numbers where ∀n≥1,∣an∣≤bn\forall n\ge 1, |a_n| \le b_n

  • (bn)(b_n) converge   ⟹  (an)\implies (a_n) converge where ∣∑n=1∞an∣≤∑n=1∞bn|\sum_{n = 1}^{\infty} a_n| \le \sum_{n = 1}^{\infty} b_n
  • (an)(a_n) diverge   ⟹  (bn)\implies (b_n) diverge

(THE ROOT TEST): Consider a sequence (an)(a_n), ∀k,ak≥0\forall k, a_k \ge 0, let l=lim sup⁡annl = \limsup \sqrt[n]{a_n}. l<1  ⟹  ∑n=1∞anl < 1\implies \sum_{n= 1}^{\infty} a_n converges, and l>1  ⟹  ∑n=1∞anl > 1 \implies \sum_{n= 1}^{\infty} a_n diverges.

A sequence is alternating if it has the form ((−1)nan)((-1)^n a_n) or ((−1)n+1an)((-1)^{n+1} a_n) where ∀n≥1,an≥0\forall n\ge 1,a_n \ge 0

(LEIBNIZ ALTERNATING SERIES TEST): Suppose that (an)n=1∞(a_n)^{\infty}_{n=1} is a monotone decreasing sequence and an≥0a_n \ge 0 where lim⁡n→∞=0\lim\limits_{n\to \infty} = 0. Then alternating series ∑n=1∞(−1)nan \sum_{n= 1}^{\infty} (-1)^n a_n converges.

(THE LIMIT COMPARISON TEST): If ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n and ∑n=1∞bn\sum\limits_{n=1}^{\infty} b_n are series with bn≥0b_n \ge 0 such that lim sup⁡n→∞∣an∣bn<∞\limsup\limits_{n\to \infty} \frac{|a_n|}{b_n} < \infty and ∑n=1∞bn<∞\sum\limits_{n=1}^{\infty} b_n < \infty, then the series ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n converges.

(THE RATIO TEST): Suppose that (an)n=1∞(a_n)^{\infty}_{n = 1} is a sequence of positive terms. If lim sup⁡n→∞an+1an<1\limsup\limits_{n\to \infty} \frac{a_{n+1}}{a_n} < 1, then ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n converges. Conversely, lim inf⁡n→∞an+1an>1\liminf\limits_{n\to \infty} \frac{a_{n+1}}{a_n} > 1 , then ∑n=1∞an\sum\limits_{n=1}^{\infty} a_n diverges.

(THE INTEGRAL TEST): Let f(x)f(x) be a positive, monotone decreasing function on [1,∞)[1,\infty). The sequence (f(n))(f(n)) is summable if and only if ∫1∞f(x)dx<∞\int^{\infty}_{1} f(x)dx < \infty.

Absolutely Convergent​

∑n=1∞an\sum_{n = 1}^{\infty} a_n is called absolutely convergent   ⟹  ∑n=1∞∣an∣\implies \sum_{n = 1}^{\infty} |a_n| converges. A series converges but is not absolutely convergent is called conditionally convergent.

  • An absolutely convergent series is convergent.

A rearrangement of a series ∑n=1∞an\sum_{n = 1}^{\infty} a_n is another series with the same terms in a different order. this can be described by a permutation π\pi of the natural numbers N\N determining the series ∑n=1∞aπ(n)\sum_{n = 1}^{\infty} a_{\pi(n)}

(THEOREM 3.3.5): Every rearrangement of an absolutely convergent series converges to the same limit.

(THEOREM 3.3.7): A rearrangement of a series ∑n=1∞an\sum_{n = 1}^{\infty} a_n is conditionally convergent series   ⟹  ∀L∈R\implies \forall L\in \R there exists a rearrangement that converges to LL