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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!
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!
