MATH 409 Lecture 24

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Lecture Slides

Course evaluations up today on Math Dept. Home Page

Tests of Convergence

divergence test
If the terms of an infinite series do not converge to zero, then the series diverges
cauchy criterion
An infinite series converges if and only if for every , there exists such that implies
comparison test (series with nonnegative terms)
Suppose for all and for large . Then convergence of the series implies convergence of while implies
integral test
Suppose that a function is positive and decreasing on then the series converges if and only if the function is improperly integrable on

Alternating Series

An infinite series is called alternating if any two neighboring terms have different signs: for all .

Leibniz Criterion

Theorem. [Alternating Series Test]. [Leibniz Criterion]. If is a decreasing sequence of positive numbers and as , then the following alternating series converges:

Proof. Let the the partial sum of order of the series. For any we have (odd terms are positive, even terms are negative)

Since the sequence is decreasing, we also have

Therefore for all . It follows that a subsequence is increasing, and a subsequence is decreasing, and both are bounded. Hence both subsequences are convergent. (but do they both converge to the same limit?)

Since as , both subsequences converge to the same limit , then is the limit of the entire sequence .

quod erat demonstrandum
Note: This theorem is not directly proved in the book. Rather it is a corollary of a more general theorem, Abel's Theorem, which is more complicated.


Examples

This series converges due to the alternating series test.

One can show that the sum is (this is a functional series in the form of a taylor series).

[Leibniz series].

After multiplying all terms by , the series satisfies all conditions of the alternating series test.

It follows that the series converges to (this is also a functional series in the form of a fourier series)

This series diverges because even though the terms decrease in absolute value, they converge to , not


Absolute Convergence of Series

An infinite series is said to converge absolutely if .

Theorem. Any absolutely convergent series is convergent.

Proof. Suppose that a series converges absolutely, that is, the series converges. By the Cauchy criterion, for every , there exists such that

for . (The outer absolute values are redundant since all terms inside will be positive anyway)

Then (by the triangle inequality) for . According to the Cauchy criterion, the series converges.

quod erat demonstrandum

Examples

The series converges due to the integral test. Since it has positive terms, it is absolutely convergent as well.

This series converges since converges absolutely by the comparison test.

This series converges due to the alternating series test, but it is not absolutely convergent as the series diverges.


Other Tests

Ratio Test (d'Alembert's Criterion)

Theorem. Let be a sequence of real numbers with for large . Suppose that the following limit exists (finite or infinite):

  1. If , then converges absolutely
  2. If , then diverges.
  3. If , the test is inconclusive. (e.g. , but converges for and diverges otherwise.)

Proof. If , then for large enough. It follows that the sequence is eventually increasing. Then Failed to parse (unknown function "\nto"): {\displaystyle a_n \nto 0} as so that the series diverges due to the divergence test.

In the case , choose some . Then for large enough. Consequently, for large enough. That is, the sequence is eventually decreasing. It follows that this sequence is bounded. Hence for some and all . Since , the geometric series converges. So does the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty C \, x^n} . By the comparison test, the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \left| a_n \right|} converges as well. The convergence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty a_n} follows from absolute convergence.

quod erat demonstrandum


Root Test

Theorem. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ a_n \right\}} be a sequence of real numbers and

  1. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r < 1} , then converges absolutely
  2. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r > 1} , then diverges.

Proof. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r > 1} , then for all . Therefore for any , there exists Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k(n) \ge n} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| a_{k(n)} \right|^{\frac{1}{k(n)}} > 1} . In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| a_{k(n)} \right| > 1} . It follows that Failed to parse (unknown function "\nto"): {\displaystyle a_k \nto 0} as so that the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^\infty a_k} diverges due to the divergence test.

In the case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r < 1} , choose some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \left( r, 1 \right)} . Then for some Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} . This implies that for all . Since , the geometric series converges. By the comparison test, the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^\infty \left| a_k \right|} converges as well. The convergence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty a_n} follows from absolute convergence.

quod erat demonstrandum


Examples

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{n}{2^n} = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \dots}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n = \frac{n}{2^n}} , then . The series converges by the ratio test.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in \mathbb{R}}

In the case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = 0} , we have a finite sum. In the case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ne 0} , let , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} \frac{\left| a_{n+1} \right|}{\left| a_n \right|} = \frac{\left| x \right|}{n+1} = 0} . Therefore the series converges absolutely for all by the ratio test.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{\left( n! \right)^2}{\left( 2n \right)!} = \frac{\left( 1! \right)^2}{2!} + \frac{\left( 2! \right)^2}{4!} + \frac{(3!)^2}{6!} + \dots}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n = \frac{\left( n! \right)^2}{\left( 2n \right)!}} , then . By the ratio test, the series converges.

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n = \left( \frac{n}{n+1} \right)^{n^2}} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{-n} = \frac{1}{\mathrm{e}}} . By the root test, the series converges.