How Do Series Converge in Normed Spaces?

In summary, series in a normed space are defined as two sequences (u_n)_n and (s_n)_n, where s_n=\sum_{k=0}^n{u_k} for every n. The elements u_n are called terms and s_n are called partial sums. A series converges if s_n\rightarrow a, otherwise it diverges. The nth term test states that a series diverges if \lim_{n\rightarrow +\infty}{u_n}\neq 0. The linearity of convergence states that if two series converge, their linear combination also converges. The deletion of finitely many terms and absolute convergence are useful tools in transforming series. In a complete norm
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Definition/Summary

In what follows, we will work in a normed space [itex](X,\|~\|)[/itex].
A series is, by definition, two sequences [itex](u_n)_n[/itex] and [itex](s_n)_n[/itex] such that [itex]s_n=\sum_{k=0}^n{u_k}[/itex] for every n.

We call the elements [itex]u_n[/itex] the terms of the series. The elements [itex]s_n[/itex] are called the partial sums. We will often denote a series by [itex]\sum_{n=0}^{+\infty}{u_n}[/itex].

We say that a series [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges to a if and only if [itex]s_n\rightarrow a[/itex]. If a series does not converge, then we say that the series diverges.

Equations



Extended explanation

Series in a normed space

For the following, we will work in a normed space [itex](X,\| ~\|)[/itex]

nth term test
If [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] is a series such that [tex]\lim_{n\rightarrow +\infty}{u_n}\neq 0,[/tex]
then the series diverges.

WARNING: The converse does not hold, i.e. if the limit does equal zero, then the series does not necessarily converge.

HINT: When given a series, always apply this test first.

Linearity of convergence
Let [itex]\lambda, \mu\in \mathbb{R}[/itex]. If [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges to u and if [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] converges to v, then the series [itex]\sum_{n=0}^{+\infty}{\lambda u_n+\mu v_n}[/itex] converges to [itex]\lambda u+\mu v[/itex].

Deletion of finitely many terms
Let [itex]p\in \mathbb{N}[/itex]. Then we have the following equivalence:
[tex]\sum_{n=0}^{+\infty}{u_n}~\text{converges iff }~\sum_{n=p}^{+\infty}{u_n}~\text{converges}[/tex]

Series in a complete normed space

In the following, we will work in a Banach space (= a complete normed space).

Cauchy criterion
A series [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges if and only if
[tex]\forall \epsilon>0:~\exists n_0:~\forall n>n_0:~\forall p:~\left\|\sum_{k=n}^{n+p}{u_k}\right\|<\epsilon[/tex]

Absolute convergence
Let [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] be a series. If the series [itex]\sum_{n=0}^{+\infty}{\|u_n\|}[/itex] converges, then the orginal series will converge. Moreover, we have

[tex]\left\|\sum_{n=0}^{+\infty}{u_n}\right\|\leq \sum_{n=0}^{+\infty}{\|u_n\|}[/tex]

NOTATION: A series such as in the above theorem is called absolutely convergent. Absolute convergence is handy because it allows you to transform a series to a series with positive real numbers.

Series with nonnegative real terms

In the following we will always work with series [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] such that all the [itex]u_n[/itex] are real and nonnegative.

Subseries
If [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] is a convergent series and if [itex]\sum_{n=0}^{+\infty}{u_{k_n}}[/itex] is a subseries, then this subseries converges. In particular, we have that

[tex]\sum_{n=0}^{+\infty}{u_{k_n}}\leq \sum_{n=0}^{+\infty}{u_n}[/tex]

Comparison test
Let [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] and [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] be two series such that [itex]u_n\leq v_n[/itex] for all n greater then a certain [itex]n_0[/itex]. Then we have:

1) If [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] converges, then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges.

If [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] diverges, then [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] diverges.

Limit comparison test
1) If [itex]\limsup_{n\rightarrow +\infty}{\frac{u_n}{v_n}}<+\infty[/itex] and if [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] converges, then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges.

2) If [itex]\liminf_{n\rightarrow +\infty}{\frac{u_n}{v_n}}>0[/itex] and if [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges, then [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] converges.

HINT: the limsup and liminf can be replaced by ordinary limits.

Comparison test 2
Let [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] and [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] be series. If there exists an m such that for every [itex]n\geq m[/itex] it holds that [itex]\frac{u_{n+1}}{u_n}\leq \frac{v_{n+1}}{v_n}[/itex], then

1) If [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] converges, then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges.

1) If [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] diverges, then [itex]\sum_{n=0}^{+\infty}{v_n}[/itex] diverges.

Cauchy condensation test
Let [itex](u_n)_n[/itex] be a nonincreasing sequence, then

[tex]\sum_{n=0}^{+\infty}{u_n}~\text{converges if and only if}~\sum_{n=0}^{+\infty}{2^nu_{2^n}}~\text{converges.}[/tex]

Cauchy's root test
Let [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] be a series. Then

1) If [itex]\limsup_{n\rightarrow +\infty}{\sqrt[n]{u_n}}<1[/itex], then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges.

2) If [itex]\limsup_{n\rightarrow +\infty}{\sqrt[n]{u_n}}>1[/itex], then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] diverges.

HINT: the limsup can be replaced by ordinary limits.

WARNING: if the limsup equals 1, then the test is inconclusive.

The ratio test of d'Alembert
Let [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] be a series. Then

1) If [itex]\limsup_{n\rightarrow +\infty}{\frac{u_{n+1}}{u_n}}<1[/itex], then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] converges.

2) If [itex]\liminf_{n\rightarrow +\infty}{\frac{u_{n+1}}{u_n}}>1[/itex], then [itex]\sum_{n=0}^{+\infty}{u_n}[/itex] diverges.

HINT: the limsup and liminf can be replaced by ordinary limits.

WARNING: if the limits equal 1, then the test is inconclusive.

The integral test
Let [itex]f:[0,+\infty[\rightarrow\mathbb{R}^+[/itex] be a nonincreasing function. Then

[tex]\sum_{n=0}^{+\infty}{f(n)}~\text{converges if and only if}~\int_1^{+\infty}{f(x)dx}<+\infty[/tex]

ADDENDUM: If [itex]f:[0,+\infty[\rightarrow\mathbb{R}^+[/itex] is a nonincreasing function, then for every [itex]n\in \mathbb{N}[/itex] holds

[tex]\sum_{k=1}^n{f(k)}\leq \int_0^n{f(x)dx}\leq \sum_{k=0}^{n-1}{f(k)}[/tex]


Series in [itex]\mathbb{R}[/itex] and [itex]\mathbb{C}[/itex]

The criterion of Dirichlet
Let [itex]\sum_{n=0}^{+\infty}{a_n}[/itex] be a (real or complex) series such that it's sequence of partial sums is bounded. Let [itex](v_n)_n[/itex] be a nonincreasing sequence of real numbers which converges to 0. Then the sequence [itex]\sum_{n=0}^{+\infty}{v_na_n}[/itex] converges.

The criterion of Abel
Let [itex]\sum_{n=0}^{+\infty}{a_n}[/itex] be a (real or complex) convergent series . Let [itex](v_n)_n[/itex] be a bounded sequence of real numbers which is either nondecreasing or nonincreasing. Then the sequence [itex]\sum_{n=0}^{+\infty}{v_na_n}[/itex] converges.

The criterion of Leibniz
Let [itex](u_n)_n[/itex] be a nonincreasing sequence of real numbers which converges to 0. Then the series [itex]\sum_{n=0}^{+\infty}{(-1)^nu_n}[/itex] converges.

ADDENDUM: Denote [itex](s_n)_n[/itex] the partial sums of the series [itex]\sum_{n=0}^{+\infty}{(-1)^nu_n}[/itex] and denote s the limit of the series. Then the sequence [itex](s_{2n})_n[/itex] is nonincreasing and [itex](s_{2n+1})_n[/itex] is nondecreasing. Moreover, we have that [itex]|s-s_n|\leq u_n[/itex].

Some special series
Geometric series
Let x be an arbitrary real or complex number. Then

[tex]\sum_{n=0}^{+\infty}{x^n}~\text{converges if and only if}~|x|<1.[/tex]

Moreover, if the series converges, then [itex]\sum_{n=0}^{+\infty}{x^n}=\frac{1}{1-x}[/itex]


p-series
Let p be a real number. Then

[tex]\sum_{n=0}^{+\infty}{\frac{1}{n^p}}~\text{converges if and only if}~p>1.[/tex]

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
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  • #3
Greg Bernhardt said:
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
Hey, @Greg Bernhardt ,
For these old Library features, you may want to place the above message at the beginning of each post/thread.

SammyS
 
  • #4
SammyS said:
For these old Library features, you may want to place the above message at the beginning of each post/thread.
You'd think it would be that easy, but then it screws up the meta description because it takes from the beginning of the first post :frown:
 
  • #5
In medias res is better. Nobody wants to read any technical preliminaries, so the comment at the end makes more sense.
 
  • #6
Greg Bernhardt said:
You'd think it would be that easy, but then it screws up the meta description because it takes from the beginning of the first post :frown:
Yup. That makes sense.

I see that you are not getting much info. regarding the authorship of these, but I'm learning a lot from @fresh_42's comments on these old Library threads and so many others.

Keep it going fresh! You too, Mark44 & others.
 

Related to How Do Series Converge in Normed Spaces?

1. What is the purpose of convergence tests for series?

Convergence tests for series are used to determine whether an infinite series converges or diverges. This is important in mathematics and science, as it helps us understand the behavior of infinite sums and their applications.

2. What are some common convergence tests for series?

Some common convergence tests for series include the ratio test, the root test, the comparison test, and the integral test. Each of these tests has its own criteria for determining convergence or divergence of a series.

3. How do you use the ratio test to test for convergence?

The ratio test involves taking the limit of the absolute value of the ratio of consecutive terms in a series. If this limit is less than 1, the series converges. If it is greater than 1, the series diverges. If the limit is exactly 1, the test is inconclusive and another test must be used.

4. Can a series converge for one convergence test but diverge for another?

Yes, it is possible for a series to converge for one convergence test but diverge for another. This is because each test has its own set of criteria and may not accurately reflect the behavior of the series for all values of the terms.

5. Are there any series which cannot be evaluated with convergence tests?

Yes, there are some series that cannot be evaluated with convergence tests. These include alternating series with terms that do not decrease in absolute value, and series with terms that do not approach zero as n approaches infinity. In these cases, other methods such as the alternating series test or the comparison test may be used.

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