Series, partial sums, convergence and the sum, divergence, and the tail series

Throughout, $\mathbb{R}$ is the complete ordered field (The real numbers) and a sequence of reals is a function $a:\mathbb{N}\to \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), written $(a_k)$; recall that $\mathbb{N}$ contains $0$.

Partial sums. Let $(a_k)$ be a sequence of reals. Its sequence of partial sums is

$s_n:={\sum }_{k<n}{a}_k(n\in \mathbb{N}),$

the finite sum of Finite sums and finite products, by recursion. In particular $s_0=0$, the empty sum, and $s_{n+1}=s_n+a_n$ for every $n$, those being exactly the two recursion clauses that define the finite sum. Note that $s_n$ is the sum of the $n$ terms $a_0,\dots ,a_{n-1}$, so the index $n$ counts terms rather than naming the last one.

Convergence, the sum, divergence. The series of $(a_k)$, written $\sum a_k$, converges when the sequence $(s_n)$ of partial sums converges (Limits and Cauchy sequences of reals), and then the sum of the series is

${\sum }_{k=0}^{\mathrm{\infty }}{a}_k:={\lim }_n{s}_n\mathrm{.}$

The series diverges when $(s_n)$ does not converge. A convergent sequence of reals has exactly one limit (A sequence has at most one limit), so the displayed symbol names a single real number and nothing further has to be checked for it to be well defined.

Series with a general starting index. Let $m\in \mathbb{N}$ and let $a$ be a function on $\{k\in \mathbb{N}:k\ge m\}$, which we call a family from $m$ and write $(a_k{)}_{k\ge m}$. The series

${\sum }_{k\ge m}{a}_k$

is by definition the series of the sequence $b_j:=a_{j+m}$, $j\in \mathbb{N}$, which is a genuine sequence of reals; it converges exactly when that series converges, and its sum is then written ${\sum }_{k=m}^{\mathrm{\infty }}{a}_k$. Its partial sums are

${\sum }_{k=m}^{n-1}{a}_k={\sum }_{j<n-m}{a}_{m+j}(n\ge m),$

in the notation of Finite sums and finite products, by recursion, the value at $n=m$ being the empty sum $0$. A sequence on $\mathbb{N}$ is the case $m=0$, and the two readings of $\sum a_k$ agree there, since $b_j=a_j$.

This clause is not a convenience. Sequences in this library are functions on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), while many of the classical series are built from expressions that are undefined at the index $0$: $1\mathrm{/}k$, $1\mathrm{/}{k}^p$ and $\mathrm{\mid }{a}_k{\mathrm{\mid }}^{1\mathrm{/}k}$ all require $k\ge 1$. Writing such a series as ${\sum }_{k\ge 1}{a}_k$ names an honest object, whereas writing it as a sequence on $\mathbb{N}$ would require a value at an index where the defining expression has none. Every statement on this page says which starting index it uses.

Tail series. For $N\in \mathbb{N}$, the $N$-th tail series of $\sum a_k$ is ${\sum }_{k\ge N}{a}_k$, that is the series of the $N$-th tail $a^{(N)}$ of Sequences of reals: bounded, eventually, frequently, tails, subsequences, whose terms are $a_j^{(N)}=a_{j+N}$. The $0$-th tail series is the series itself.

• "Diverges" here means "does not converge", and nothing more. A divergent series may have partial sums that run away to $+\mathrm{\infty }$, or to $-\mathrm{\infty }$, or that oscillate without settling anywhere. The three behaviours are not distinguished by the word, and no statement on this page uses "diverges" to mean "the partial sums are unbounded" unless it says so.

• The symbol ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k$ is defined only for a convergent series. It denotes a real number, not a formal object, and it is illegitimate to write it down before convergence has been established. Where a proof needs to speak of the series without knowing whether it converges, it speaks of $(a_k)$ and of $(s_n)$.

• Two indices, doing different work. The index $k$ runs over the terms and is bound; the index $n$ runs over the partial sums and is the variable in which the limit is taken. Confusing them is the commonest slip in the subject, and it is the reason the definition above fixes $s_n={\sum }_{k<n}{a}_k$ rather than ${\sum }_{k\le n}{a}_k$: with this choice the recursion $s_{n+1}=s_n+a_n$ is the one supplied by Finite sums and finite products, by recursion, with no shift anywhere.