4. Series

Def.: for some sequence

Sequence of partial sums:

Absolutely convergent: converges Conditionally convergent: converges but does not

Necessary Condition for Convergence

Lemma

If a series converges, then

Not sufficient: converges, but its corresponding series does not. Intuitively: terms shrink, but not fast enough to limit the sum. Reason: We can group the terms:

Riemann Rearrangement Theorem

Theorem

If is conditionally convergent and is any real number, then there exists a bijection s.t.

Proof:

  • For it to be conditionally convergent, must contain infinitely many positive and negative terms (an inf. large pool).
  • We first add positive terms until we overshoot . We then add negative terms until we undershoot . We repeat this process into infinity.
  • We over/undershoot by at most the size of the last term. But since , both and . Hence, we overshoot and undershoot ever closer to , eventually converging.

Absolutely convergent series: This does not apply! We don’t have the infinite pools. By the Absolute Series Rearrangement Theorem,

Leibnitz Criterion

Theorem

Let be a monotonically decreasing sequence of non-negative real numbers converging to zero.

Then the alternating series converges, and:

Proof:

  • Even: ( mono decreasing)
  • Odd:

Cauchy Criterion

Partial sums are sequences can be [[3. Sequences#3. Sequences#Cauchy Sequence|Cauchy Sequence]] converge iff Cauchy Sequence

We can simplify the def since the terms cancel:

Absolute Convergence Implies Convergence

Theorem

Every absolutely convergent series converges, and the generalized triangle inequality holds:

Quotient Test (Quotienten-Kriterium)

Theorem

. Then:

  • If , converges absolutely.
  • If , does not converge.
  • If , the test is inconclusive

Root Test (Wurzelkriterium)

Same as Quotient Test, but with

Cauchy Product

Theorem

Let and be absolutely converging series. Then:

The series to the right converges absolutely.

Power Series

Def:

Radius of convergence: The s.t.:

  • the series converges for with and
  • the series diverges for with .

Interval of convergence:

Lemma

The radius of convergence can be computed with the following formula: