Summation

The summation symbol

Mathematical notation uses a symbol that compactly represents summation of many similar terms: the summation symbol, ${\displaystyle \textstyle \sum }$, an enlarged form of the upright capital Greek letter Sigma. This is defined as:

${\displaystyle \sum _{i\mathop {=} m}^{n}a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n}}$

where i represents the index of summation; a_i is an indexed variable representing each successive term in the series; m is the lower bound of summation, and n is the upper bound of summation. The "i = m" under the summation symbol means that the index i starts out equal to m. The index, i, is incremented by 1 for each successive term, stopping when i = n.[lower-alpha 2]

Here is an example showing the summation of squares:

${\displaystyle \sum _{i=3}^{6}i^{2}=3^{2}+4^{2}+5^{2}+6^{2}=86.}$

Informal writing sometimes omits the definition of the index and bounds of summation when these are clear from context, as in:

${\displaystyle \sum a_{i}^{2}=\sum _{i\mathop {=} 1}^{n}a_{i}^{2}.}$

One often sees generalizations of this notation in which an arbitrary logical condition is supplied, and the sum is intended to be taken over all values satisfying the condition. Here are some common examples:

${\displaystyle \sum _{0\leq k<100}f(k)}$

is the sum of ${\displaystyle f(k)}$ over all (integers) ${\displaystyle k}$ in the specified range,

${\displaystyle \sum _{x\mathop {\in } S}f(x)}$

is the sum of ${\displaystyle f(x)}$ over all elements ${\displaystyle x}$ in the set ${\displaystyle S}$, and

${\displaystyle \sum _{d|n}\;\mu (d)}$

is the sum of ${\displaystyle \mu (d)}$ over all positive integers ${\displaystyle d}$ dividing ${\displaystyle n}$.[lower-alpha 3]

There are also ways to generalize the use of many sigma signs. For example,

${\displaystyle \sum _{i,j}}$

is the same as

${\displaystyle \sum _{i}\sum _{j}.}$

A similar notation is applied when it comes to denoting the product of a sequence, which is similar to its summation, but which uses the multiplication operation instead of addition (and gives 1 for an empty sequence instead of 0). The same basic structure is used, with ${\displaystyle \textstyle \prod }$, an enlarged form of the Greek capital letter pi, replacing the ${\displaystyle \textstyle \sum }$.

It is possible to sum fewer than 2 numbers:

• If the summation has one summand ${\displaystyle x}$, then the evaluated sum is ${\displaystyle x}$.
• If the summation has no summands, then the evaluated sum is zero, because zero is the identity for addition. This is known as the empty sum.

These degenerate cases are usually only used when the summation notation gives a degenerate result in a special case. For example, if ${\displaystyle n=m}$ in the definition above, then there is only one term in the sum; if ${\displaystyle n=m-1}$, then there is none.

${\displaystyle \sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)\quad }$ (distributivity)

${\displaystyle \sum _{n=s}^{t}f(n)\pm \sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left(f(n)\pm g(n)\right)\quad }$ (commutativity and associativity)

${\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+p}^{t+p}f(n-p)\quad }$ (index shift)

${\displaystyle \sum _{n\in B}f(n)=\sum _{m\in A}f(\sigma (m)),\quad }$ for a bijection σ from a finite set A onto a set B (index change); this generalizes the preceding formula.

${\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s}^{j}f(n)+\sum _{n=j+1}^{t}f(n)\quad }$ (splitting a sum, using associativity)

${\displaystyle \sum _{n=a}^{b}f(n)=\sum _{n=0}^{b}f(n)-\sum _{n=0}^{a-1}f(n)\quad }$ (a variant of the preceding formula)

${\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=0}^{t-s}f(t-n)\quad }$ (the sum from the first term up to the last is equal to the sum from the last down to the first)

${\displaystyle \sum _{n=0}^{t}f(n)=\sum _{n=0}^{t}f(t-n)\quad }$ (a particular case of the formula above)

${\displaystyle \sum _{i=k_{0}}^{k_{1}}\sum _{j=l_{0}}^{l_{1}}a_{i,j}=\sum _{j=l_{0}}^{l_{1}}\sum _{i=k_{0}}^{k_{1}}a_{i,j}\quad }$ (commutativity and associativity, again)

${\displaystyle \sum _{k\leq j\leq i\leq n}a_{i,j}=\sum _{i=k}^{n}\sum _{j=k}^{i}a_{i,j}=\sum _{j=k}^{n}\sum _{i=j}^{n}a_{i,j}=\sum _{j=0}^{n-k}\sum _{i=k}^{n-j}a_{i+j,i}\quad }$ (another application of commutativity and associativity)

${\displaystyle \sum _{n=2s}^{2t+1}f(n)=\sum _{n=s}^{t}f(2n)+\sum _{n=s}^{t}f(2n+1)\quad }$ (splitting a sum into its odd and even parts, for even indexes)

${\displaystyle \sum _{n=2s+1}^{2t}f(n)=\sum _{n=s+1}^{t}f(2n)+\sum _{n=s+1}^{t}f(2n-1)\quad }$ (splitting a sum into its odd and even parts, for odd indexes)

${\displaystyle \left(\sum _{i=0}^{n}a_{i}\right)\left(\sum _{j=0}^{n}b_{j}\right)=\sum _{i=0}^{n}\sum _{j=0}^{n}a_{i}b_{j}\quad }$ (distributivity)

${\displaystyle \sum _{i=s}^{m}\sum _{j=t}^{n}{a_{i}}{c_{j}}=\left(\sum _{i=s}^{m}a_{i}\right)\left(\sum _{j=t}^{n}c_{j}\right)\quad }$ (distributivity allows factorization)

${\displaystyle \sum _{n=s}^{t}\log _{b}f(n)=\log _{b}\prod _{n=s}^{t}f(n)\quad }$ (the logarithm of a product is the sum of the logarithms of the factors)

${\displaystyle C^{\sum \limits _{n=s}^{t}f(n)}=\prod _{n=s}^{t}C^{f(n)}\quad }$ (the exponential of a sum is the product of the exponential of the summands)

${\displaystyle \sum _{i=1}^{n}c=nc\quad }$ for every c that does not depend on i

${\displaystyle \sum _{i=0}^{n}i=\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}\qquad }$ (Sum of the simplest arithmetic progression, consisting of the n first natural numbers.)[2]

${\displaystyle \sum _{i=1}^{n}(2i-1)=n^{2}\qquad }$ (Sum of first odd natural numbers)

${\displaystyle \sum _{i=0}^{n}2i=n(n+1)\qquad }$ (Sum of first even natural numbers)

${\displaystyle \sum _{i=1}^{n}\log i=\log n!\qquad }$ (A sum of logarithms is the logarithm of the product)

${\displaystyle \sum _{i=0}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\qquad }$ (Sum of the first squares, see square pyramidal number.) [2]

${\displaystyle \sum _{i=0}^{n}i^{3}=\left(\sum _{i=0}^{n}i\right)^{2}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}\qquad }$ (Nicomachus's theorem) [2]

More generally,

${\displaystyle \sum _{i=0}^{n}i^{p}={\frac {(n+1)^{p+1}}{p+1}}+\sum _{k=1}^{p}{\frac {B_{k}}{p-k+1}}{p \choose k}(n+1)^{p-k+1},}$

where ${\displaystyle B_{k}}$ denotes a Bernoulli number (that is Faulhaber's formula).

In the following summations, a is assumed to be different from 1.

${\displaystyle \sum _{i=0}^{n-1}a^{i}={\frac {1-a^{n}}{1-a}}}$ (sum of a geometric progression)

${\displaystyle \sum _{i=0}^{n-1}{\frac {1}{2^{i}}}=2-{\frac {1}{2^{n-1}}}}$ (special case for a = 1/2)

${\displaystyle \sum _{i=0}^{n-1}ia^{i}={\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}}}$ (a times the derivative with respect to a of the geometric progression)

${\displaystyle {\begin{aligned}\sum _{i=0}^{n-1}\left(b+id\right)a^{i}&=b\sum _{i=0}^{n-1}a^{i}+d\sum _{i=0}^{n-1}ia^{i}\\&=b\left({\frac {1-a^{n}}{1-a}}\right)+d\left({\frac {a-na^{n}+(n-1)a^{n+1}}{(1-a)^{2}}}\right)\\&={\frac {b(1-a^{n})-(n-1)da^{n}}{1-a}}+{\frac {da(1-a^{n-1})}{(1-a)^{2}}}\end{aligned}}}$

(sum of an arithmetico–geometric sequence)

There exist very many summation identities involving binomial coefficients (a whole chapter of Concrete Mathematics is devoted to just the basic techniques). Some of the most basic ones are the following.

${\displaystyle \sum _{i=0}^{n}{n \choose i}a^{n-i}b^{i}=(a+b)^{n},}$ the binomial theorem

${\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n},}$ the special case where a = b = 1

${\displaystyle \sum _{i=0}^{n}{n \choose i}p^{i}(1-p)^{n-i}=1}$, the special case where p = a = 1 – b, which, for ${\displaystyle 0\leq p\leq 1,}$ expresses the sum of the binomial distribution

${\displaystyle \sum _{i=0}^{n}i{n \choose i}=n(2^{n-1}),}$ the value at a = b = 1 of the derivative with respect to a of the binomial theorem

${\displaystyle \sum _{i=0}^{n}{\frac {n \choose i}{i+1}}={\frac {2^{n+1}-1}{n+1}},}$ the value at a = b = 1 of the antiderivative with respect to a of the binomial theorem

In the following summations, ${\displaystyle {}_{n}P_{k}}$ is the number of k-permutations of n.

${\displaystyle \sum _{i=0}^{n}{}_{i}P_{k}{n \choose i}={}_{n}P_{k}(2^{n-k})}$

${\displaystyle \sum _{i=1}^{n}{}_{i+k}P_{k+1}=\sum _{i=1}^{n}\prod _{j=0}^{k}(i+j)={\frac {(n+k+1)!}{(n-1)!(k+2)}}}$

${\displaystyle \sum _{i=0}^{n}i!\cdot {n \choose i}=\sum _{i=0}^{n}{}_{n}P_{i}=\lfloor n!\cdot e\rfloor ,\quad n\in \mathbb {Z} ^{+}}$, where and ${\displaystyle \lfloor x\rfloor }$ denotes the floor function.

${\displaystyle \sum _{k=0}^{m}\left({\begin{array}{c}n+k\\n\\\end{array}}\right)=\left({\begin{array}{c}n+m+1\\n+1\\\end{array}}\right)}$

${\displaystyle \sum _{i=k}^{n}{i \choose k}={n+1 \choose k+1}}$

${\displaystyle \sum _{i=0}^{n}i\cdot i!=(n+1)!-1}$

${\displaystyle \sum _{i=0}^{n}{m+i-1 \choose i}={m+n \choose n}}$

${\displaystyle \sum _{i=0}^{n}{n \choose i}^{2}={2n \choose n}}$

${\displaystyle \sum _{i=1}^{n}{\frac {1}{i}}=H_{n}}$ (that is the nth harmonic number)

${\displaystyle \sum _{i=1}^{n}{\frac {1}{i^{k}}}=H_{n}^{k}}$ (that is a generalized harmonic number)