Sequences and series provide the convergence language behind many time-series representations. A sequence describes an ordered set of terms, while a series asks whether the cumulative effect of those terms approaches a finite limit.
This becomes important when a dynamic model is expanded into an infinite history of shocks or observations. Conditions such as geometric decay, absolute summability, and mean-square convergence determine whether those infinite representations are mathematically stable and can be approximated accurately with a finite number of terms.
Worked calculation: a geometric series
Consider the terms
$$ 1,\quad0.5,\quad0.5^2,\quad0.5^3,\ldots $$
The partial sum through the fourth power is
$$ S_4=1+0.5+0.25+0.125+0.0625=1.9375. $$
The infinite sum is
$$ \sum_{j=0}^{\infty}0.5^j=\frac{1}{1-0.5}=2. $$
So the omitted tail after $S_4$ is
$$ 2-1.9375=0.0625. $$
This is why a stable linear time-series representation can be approximated by finitely many terms: when the weights decay geometrically, the discarded tail has a known bound.
The figure shows the partial sums approaching 2. Each added term improves the approximation, but the size of the improvement shrinks because the geometric weights decay.
Sequences
A sequence is an ordered list of numbers, formally a function that maps an index such as a natural number $n$ to a value $a_n$:
$$ a_1,a_2,a_3,\dots,a_n,\dots $$
A sequence converges to $a$ if its terms become arbitrarily close to $a$ as $n$ increases:
$$ \lim_{n\to\infty}a_n=a. $$
Equivalently, for every $\epsilon>0$, there is an index $N$ such that
$$ |a_n-a|<\epsilon $$
for all $n>N$. If no finite limit satisfies this condition, the sequence diverges.
Examples of Sequences
A convergent example is
$$ a_n=\frac{n}{n+2}. $$
Its first terms are
$$ \frac13,\frac24,\frac35,\dots, $$
and
$$ \lim_{n\to\infty}\frac{n}{n+2}=1. $$
By contrast,
$$ a_n=4^n $$
grows without bound, so it diverges. The sequence
$$ a_n=n+1 $$
also diverges because its terms increase indefinitely.
Another convergent example is
$$ a_n=\frac{1}{n^3}, $$
for which
$$ \lim_{n\to\infty}\frac{1}{n^3}=0. $$
These examples concern the behavior of the individual terms. A series asks a different question: what happens when the terms are accumulated?
Partial Sums
For a sequence ${a_n}$, the partial sum of the first $n$ terms is
$$ s_n=a_1+a_2+\dots+a_n. $$
Thus,
$$ s_1=a_1, \qquad s_2=a_1+a_2, \qquad s_3=a_1+a_2+a_3, $$
and so on. Infinite-series convergence is defined through the behavior of this sequence of partial sums.
Series
A series is the formal sum of the terms of a sequence. If the partial sums converge to a finite limit $s$, then
$$ \sum_{k=1}^{\infty}a_k =\lim_{n\to\infty}s_n =s. $$
If the partial sums do not approach a finite limit, the series diverges. Therefore, a term sequence approaching zero is necessary for series convergence but is not sufficient; the harmonic series later in the chapter is the standard counterexample.
Geometric Series and Rational Functions
A geometric series has the form
$$ \sum_{k=0}^{\infty}r^k. $$
When $|r|<1$,
$$ \sum_{k=0}^{\infty}r^k=\frac{1}{1-r}. $$
Equivalently,
$$ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k, \qquad |x|<1. $$
This identity is especially useful in time-series analysis because inverse lag polynomials can often be expanded as geometric or power series. The convergence condition determines whether the resulting infinite-lag representation is stable.
Examples of Convergent Series
A geometric series with ratio $1/3$ is
$$ \sum_{k=0}^{\infty}\frac{1}{3^k} =\frac{1}{1-1/3} =\frac32. $$
A p-series is
$$ \sum_{k=1}^{\infty}\frac{1}{k^p}. $$
It converges for $p>1$. In particular,
$$ \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}. $$
An alternating series changes sign from term to term. The alternating harmonic series,
$$ \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}, $$
converges to
$$ \ln 2. $$
These examples illustrate different reasons for convergence: geometric decay, sufficiently fast polynomial decay, and cancellation in an alternating sequence.
Examples of Divergent Series
A geometric series with ratio larger than 1 in magnitude diverges. For example,
$$ \sum_{k=1}^{\infty}4^k=4+16+64+\dots $$
has partial sums that grow without bound.
An arithmetic-term series such as
$$ \sum_{k=1}^{\infty}(2k+3)=5+7+9+\dots $$
also diverges because the terms themselves do not approach zero.
The harmonic series
$$ \sum_{k=1}^{\infty}\frac{1}{k} =1+\frac12+\frac13+\dots $$
is more subtle. Its terms do approach zero, but the partial sums still grow without bound. This is why checking $a_n\to0$ is only a necessary condition for convergence.
Absolute Convergence
A series is absolutely convergent if
$$ \sum_{k=1}^{\infty}|a_k| $$
converges. Absolute convergence implies ordinary convergence. The converse is not always true: the alternating harmonic series converges, but the corresponding series of absolute values is the divergent harmonic series.
Absolute summability is particularly useful for linear filters because it gives a strong form of stability and makes many rearrangements and expectation calculations straightforward.
Convergence Tests
Different series have different structures, so no single convergence test is best in every case. The tests below give common ways to compare the tail behavior of a series with a known benchmark.
Integral test. Suppose $f(x)$ is positive, continuous, and decreasing for $x\ge1$, with $f(n)=a_n$. Then
$$ \sum_{n=1}^{\infty}a_n $$
and
$$ \int_1^{\infty}f(x)\,dx $$
converge or diverge together. For example, $\sum 1/n^2$ converges because
$$ \int_1^{\infty}\frac{1}{x^2}\,dx=1. $$
Comparison test. If
$$ 0\le a_n\le b_n $$
and $\sum b_n$ converges, then $\sum a_n$ also converges. For example,
$$ \frac{1}{n^3+2}<\frac{1}{n^3}, $$
and $\sum1/n^3$ is a convergent p-series, so
$$ \sum_{n=1}^{\infty}\frac{1}{n^3+2} $$
converges.
Limit comparison test. For positive-term series, if
$$ \lim_{n\to\infty}\frac{a_n}{b_n}=c, \qquad 0<c<\infty, $$
then $\sum a_n$ and $\sum b_n$ have the same convergence behavior. For
$$ a_n=\frac{n^2+1}{n^4+3}, \qquad b_n=\frac{1}{n^2}, $$
we have
$$ \lim_{n\to\infty} \frac{\frac{n^2+1}{n^4+3}}{\frac1{n^2}} = \lim_{n\to\infty}\frac{n^4+n^2}{n^4+3} =1. $$
Because $\sum1/n^2$ converges, so does the original series.
Alternating series test. For an alternating series
$$ \sum(-1)^na_n, $$
if $a_n\ge0$ decreases eventually and $a_n\to0$, then the series converges. The alternating harmonic series satisfies these conditions.
Ratio test. Let
$$ L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|. $$
If $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. For
$$ a_n=\frac{n!}{n^n}, $$
$$ \frac{a_{n+1}}{a_n} =\left(\frac{n}{n+1}\right)^n, $$
so
$$ L=e^{-1}<1. $$
Therefore the series converges.
Root test. Let
$$ L=\limsup_{n\to\infty}|a_n|^{1/n}. $$
If $L<1$, the series converges absolutely; if $L>1$, it diverges. For
$$ a_n=\left(\frac34\right)^n, $$
$$ L=\frac34<1, $$
so the geometric series converges.
Mean-Square Convergence
For random variables, convergence can be defined through expected squared error. A sequence $X_n$ converges to $X$ in mean square if
$$ E[(X_n-X)^2]\to0 \quad\text{as }n\to\infty. $$
This notion is important for stochastic processes because infinite linear representations are often interpreted as limits of finite random sums.
Inverting the MA(1) Model
Consider
$$ X_t=Z_t+\beta Z_{t-1}, $$
where $Z_t$ is white noise with mean zero and variance $\sigma_Z^2$. Rearranging and repeatedly substituting gives
$$ Z_t=\sum_{k=0}^{\infty}(-\beta)^kX_{t-k}. $$
This inverse representation is stable when the geometric coefficients decay, which for MA(1) means
$$ |\beta|<1. $$
Autocovariance Function of the MA(1) Process
The autocovariance function is
$$ \gamma(k)= \begin{cases} (1+\beta^2)\sigma_Z^2, & k=0,\\ \beta\sigma_Z^2, & |k|=1,\\ 0, & |k|>1. \end{cases} $$
The finite covariance range reflects the finite shock duration of the MA(1): observations more than one period apart share no common innovation.
Series Convergence in Mean-Square Sense
Consider the partial inverse
$$ S_n=\sum_{k=0}^{n}(-\beta)^kX_{t-k}. $$
Repeated substitution gives the exact decomposition
$$ Z_t =S_n+(-\beta)^{n+1}Z_{t-n-1}. $$
Therefore the truncation error is
$$ Z_t-S_n=(-\beta)^{n+1}Z_{t-n-1}, $$
and its mean square is
$$ E[(Z_t-S_n)^2] =\beta^{2(n+1)}\sigma_Z^2. $$
Condition for Convergence
Mean-square convergence requires
$$ \beta^{2(n+1)}\sigma_Z^2\to0, $$
which occurs exactly when
$$ |\beta|<1. $$
Thus the same condition that makes the inverse weights decay also makes the finite truncations converge to the innovation in mean square.
Invertibility Condition
The MA polynomial is
$$ \beta(B)=1+\beta B. $$
Its zero is
$$ B=-\frac{1}{\beta}. $$
Requiring that zero to lie outside the unit circle gives
$$ \left|-\frac{1}{\beta}\right|>1 \quad\Longleftrightarrow\quad |\beta|<1. $$
Invertibility therefore links a polynomial root condition, geometric-series convergence, and stable recovery of the innovations.
Student guide: why convergence appears in time-series filters
Infinite series appear whenever a stable dynamic model is rewritten as a weighted history of shocks or observations. The geometric identity
$$ \sum_{j=0}^{\infty}r^j=\frac{1}{1-r}, \qquad |r|<1, $$
is the simplest example.
With $r=0.5$,
$$ S_4=1+0.5+0.25+0.125+0.0625=1.9375, \qquad S_\infty=2. $$
The omitted tail is
$$ \sum_{j=5}^{\infty}0.5^j =\frac{0.5^5}{1-0.5}=0.0625. $$
For a linear process
$$ X_t=\sum_{j=0}^{\infty}\psi_j\varepsilon_{t-j}, $$
square summability,
$$ \sum_{j=0}^{\infty}\psi_j^2<\infty, $$
ensures a finite second moment when the shocks are uncorrelated with finite variance. If $\psi_j=0.5^j$, then
$$ \sum_{j=0}^{\infty}0.25^j =\frac{1}{0.75} =\frac43. $$
Thus, if $\mathrm{Var}(\varepsilon_t)=\sigma^2$,
$$ \mathrm{Var}(X_t)=\frac43\sigma^2. $$
This calculation connects numerical series convergence directly to the variance of a stochastic linear filter.
Numerical stability
Finite computations truncate infinite representations. For a geometric series with $|r|<1$, the absolute tail after retaining terms through $m$ satisfies
$$ \left|\sum_{j=m+1}^{\infty}r^j\right| =\frac{|r|^{m+1}}{|1-r|} $$
for positive $r$, and is bounded by
$$ \frac{|r|^{m+1}}{1-|r|} $$
in general.
When $r=0.99$, truncation error decays far more slowly than when $r=0.5$. This is the numerical counterpart of persistence in a near-unit-root time-series model: a formally convergent representation can still require a long history before finite approximations become accurate.