How Do Series Converge in Normed Spaces?

[Greg Bernhardt]

Definition/Summary

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

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

We say that a series $\sum_{n=0}^{+\infty}{u_n}$ converges to a if and only if $s_n\rightarrow a$. 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 $(X,\| ~\|)$

nth term test
If $\sum_{n=0}^{+\infty}{u_n}$ is a series such that $$\lim_{n\rightarrow +\infty}{u_n}\neq 0,$$
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 $\lambda, \mu\in \mathbb{R}$. If $\sum_{n=0}^{+\infty}{u_n}$ converges to u and if $\sum_{n=0}^{+\infty}{v_n}$ converges to v, then the series $\sum_{n=0}^{+\infty}{\lambda u_n+\mu v_n}$ converges to $\lambda u+\mu v$.

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

Series in a complete normed space

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

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

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

$$\left\|\sum_{n=0}^{+\infty}{u_n}\right\|\leq \sum_{n=0}^{+\infty}{\|u_n\|}$$

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 $\sum_{n=0}^{+\infty}{u_n}$ such that all the $u_n$ are real and nonnegative.

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

$$\sum_{n=0}^{+\infty}{u_{k_n}}\leq \sum_{n=0}^{+\infty}{u_n}$$

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

1) If $\sum_{n=0}^{+\infty}{v_n}$ converges, then $\sum_{n=0}^{+\infty}{u_n}$ converges.

If $\sum_{n=0}^{+\infty}{u_n}$ diverges, then $\sum_{n=0}^{+\infty}{v_n}$ diverges.

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

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

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

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

1) If $\sum_{n=0}^{+\infty}{v_n}$ converges, then $\sum_{n=0}^{+\infty}{u_n}$ converges.

1) If $\sum_{n=0}^{+\infty}{u_n}$ diverges, then $\sum_{n=0}^{+\infty}{v_n}$ diverges.

Cauchy condensation test
Let $(u_n)_n$ be a nonincreasing sequence, then

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

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

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

2) If $\limsup_{n\rightarrow +\infty}{\sqrt[n]{u_n}}>1$, then $\sum_{n=0}^{+\infty}{u_n}$ 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 $\sum_{n=0}^{+\infty}{u_n}$ be a series. Then

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

2) If $\liminf_{n\rightarrow +\infty}{\frac{u_{n+1}}{u_n}}>1$, then $\sum_{n=0}^{+\infty}{u_n}$ 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 $f:[0,+\infty[\rightarrow\mathbb{R}^+$ be a nonincreasing function. Then

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

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

$$\sum_{k=1}^n{f(k)}\leq \int_0^n{f(x)dx}\leq \sum_{k=0}^{n-1}{f(k)}$$


Series in $\mathbb{R}$ and $\mathbb{C}$

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

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

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

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

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

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

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


p-series
Let p be a real number. Then

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

* 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!

 

[fresh_42]

Here's a proof of Cauchy's condensation test:
https://www.physicsforums.com/threads/basic-math-challenge-september-2018.954490/#post-6051199

 

[SammyS]

* 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

 

[Greg Bernhardt]

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

 

[fresh_42]

In medias res is better. Nobody wants to read any technical preliminaries, so the comment at the end makes more sense.

 

[SammyS]

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

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.