The sum of a sequence of terms is called a series.
Let \((a_n)_{n\geq k}\) be a sequence starting with index \(k\). Now sum over the first \(n-k+1\) terms:
\[\begin{array}{rl} s_n =& \sum\limits_{i=k}^n a_i\\ =& \underbrace{\underbrace{\underbrace{\underbrace{a_k}_{s_k} + a_{k+1}}_{s_{k+1}}+a_{k+2}}_{s_{k+2}}+...+a_n}_{s_n} \end{array}\]
The sequence \((s_n) = (s_k, s_{k+1}, s_{k+2}, ...)\) is called an infinite series \(\sum\limits_{i=1}^\infty a_i\). If \((s_n)\) converges, its limit is called \(\sum\limits_{i=k}^\infty a_i\).
We can state that if \(\sum\limits_{i=k}^\infty a_i\) converges \(\Rightarrow (a_i)_{n\geq k}\) is a null sequence. \(\Leftarrow\) does not hold, see Harmonic series.
Examples
- \(\sum\limits_{i=1}^\infty i = 1+2+3+...\) diverges
- \(\sum\limits_{i=1}^\infty (-1)^i\) diverges, \(s_n = \sum\limits_{i=1}^n(-1)^i = \begin{cases} 0, & \text{if } n \text{ even} \\ -1, & \text{if } n \text{ odd} \end{cases}\)
- Geometric Series: \(\sum\limits_{i=0}^\infty q^i \begin{cases} |q|<1: \text{ converges to }\frac{1}{1-q} \\ |q|\geq 1: \text{ diverges} \end{cases}\)
- Leibnitz-Series: \(\sum\limits_{i=0}^\infty(-1)^i\frac{1}{2i+1}\) converges to \(\frac{\pi}{4}\)
- Alternating Harmonic series: \(\sum\limits_{i=0}^\infty(-1)^i\frac{1}{i+1}\) converges to \(\ln 2\)
- \(\sum\limits_{i=0}^\infty 2^{-i}\) converges to \(2\) and \(\sum\limits_{i=1}^\infty 2^{-i}=1\).
Linearity of Summation
Before deriving closed forms for specific series, it is worth recording a property used throughout: summation distributes over addition and scalar multiplication. For a constant \(a\), a constant factor \(b\), and any sequence \((x_i)\),
\[\sum_{i=k}^n(a+bx_i) = a(n-k+1)+b\sum_{i=k}^n x_i.\]
This follows by splitting the sum term by term: each of the \(n-k+1\) summands contributes its own copy of \(a\), which adds up to \(a(n-k+1)\), while the \(bx_i\) parts add up to \(b\) times \(\sum x_i\), since a common factor can be pulled out of every term. In particular, taking \(b=0\) shows \(\sum_{i=k}^n a=a(n-k+1)\) (summing a constant \(n-k+1\) times), and taking \(a=0\) shows \(\sum_{i=k}^n bx_i=b\sum_{i=k}^n x_i\). For the common case \(k=1\), this reads \(\sum_{i=1}^n(a+bx_i)=an+b\sum_{i=1}^n x_i\).
Partial Sums of a Geometric Series
The \(n\)-th partial sum of a geometric series can be brought into closed form. Starting from
\[s_n = \sum\limits_{i=0}^n a_0q^i = a_0+a_0q+a_0q^2+\cdots+a_0q^n = a_0\left(1+q+q^2+\cdots+q^n\right),\tag{1}\]
multiply through by \(q\):
\[qs_n = a_0\left(q+q^2+q^3+\cdots+q^{n+1}\right).\tag{2}\]
Subtracting \((2)\) from \((1)\), every middle term cancels, leaving only the first and the last:
\[s_n-qs_n = a_0\left(1-q^{n+1}\right).\]
Factoring \(s_n\) on the left and, for \(q\neq 1\), dividing by \(1-q\) gives the closed form
\[\boxed{s_n = a_0\frac{1-q^{n+1}}{1-q}}\qquad(q\neq 1).\]
This matches the infinite-sum limit above: for \(|q|<1\), \(q^{n+1}\to 0\) as \(n\to\infty\), so \(s_n\to \frac{a_0}{1-q}\).
Arithmetic Series
An arithmetic sequence increases (or decreases) by a fixed amount \(d\) at each step, \(a_i=a_0+id\), and summing its first \(n+1\) terms gives an arithmetic series. The simplest and most famous case is \(d=1\), \(a_0=0\): the sum of the first \(n\) positive integers,
\[S=\sum_{i=0}^n i = 0+1+2+\cdots+n.\]
Writing the same sum forwards and backwards and adding term by term pairs up every term with its complement to \(n\):
\[ \begin{array}{rcccccccc} S &=& 0 &+& 1 &+& \cdots &+& n\\ S &=& n &+& (n-1) &+& \cdots &+& 0\\ \hline 2S &=& n &+& n &+& \cdots &+& n \end{array} \]
Every one of the \(n+1\) columns adds up to exactly \(n\), so \(2S=n(n+1)\), giving the closed form usually attributed to Gauss:
\[\boxed{\sum_{i=0}^n i = \frac{n(n+1)}{2}}.\]
For a general arithmetic sequence \(a_i=a_0+id\), the linearity of summation reduces the sum to this special case:
\[ \sum_{i=0}^n a_i=\sum_{i=0}^n(a_0+id)=a_0(n+1)+d\sum_{i=0}^n i=a_0(n+1)+d\frac{n(n+1)}{2}=(n+1)\left(a_0+\frac{dn}{2}\right). \]
Since \(a_0+\frac{dn}{2}\) is the average of the first term \(a_0\) and the last term \(a_n=a_0+dn\), this says the sum of an arithmetic series is always the number of terms times the average of its first and last term, \((n+1)\cdot\frac{a_0+a_n}{2}\).
Visual Proofs
Some geometric series can be seen to converge, and to what value, directly from a picture, without invoking the closed form above at all.
\(\sum_{n=1}^\infty\frac{1}{2^n}=1\)
Take a square of area \(1\) and repeatedly cut the remaining piece in half, alternating a vertical and a horizontal cut:
Each cut removes exactly half of what remained, so after \(n\) cuts the shaded pieces have areas \(\frac12,\frac14,\dots,\frac{1}{2^n}\), and the still-unshaded square left over has area \(\frac{1}{2^n}\) too — the same size as the piece just cut off, since halving it again would produce that piece plus an equally sized remainder. Shrinking the leftover square to a point as \(n\to\infty\) shows the shaded pieces exhaust the entire unit square, so
\[\sum_{n=1}^{\infty}\frac{1}{2^n}=\frac12+\frac14+\frac18+\frac1{16}+\cdots=1.\]
\(\sum_{n=1}^\infty\frac{1}{4^n}=\frac13\)
Take a square of area \(1\) and split it into four equal quadrants. Shade the bottom-left quadrant (area \(\frac14\)), and repeat the same construction inside the top-right quadrant, forever:
Write \(S\) for the total shaded area. At every level of the recursion, exactly one of the four congruent sub-squares is shaded, one keeps recursing, and the remaining two are left permanently blank — they are never touched again at any later level. So for every shaded square of some area \(A\) there are exactly two blank squares of that same area \(A\) that will never be shaded. Summing over all levels, the blank area is exactly \(2S\), and together with the shaded area \(S\) itself this accounts for the whole square:
\[S+2S=1\qquad\Longrightarrow\qquad S=\frac13.\]
Since the shaded squares have areas \(\frac14,\frac1{16},\frac1{64},\ldots\), this proves
\[\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac14+\frac1{16}+\frac1{64}+\cdots=\frac13,\]
which also matches the closed form derived above, \(\frac{a_0}{1-q}\) with \(a_0=q=\frac14\).
Application: Exponential Growth Models
A population (or any quantity) that changes by a fixed proportion \(r\) at each step follows the recursion
\[P_{i+1}=(1+r)P_i,\]
meaning each new value is the previous one plus an \(r\)-fraction of it (growth for \(r>0\), decay for \(-1<r<0\)). Unrolling the recursion from \(P_0\),
\[P_1=(1+r)P_0,\quad P_2=(1+r)P_1=(1+r)^2P_0,\quad\ldots,\quad P_n=(1+r)^nP_0,\]
shows that \((P_i)\) is itself a geometric sequence with ratio \(q=1+r\). Its partial sums therefore follow directly from the closed form derived above, with \(a_0\to P_0\) and \(q\to1+r\):
\[\sum_{i=0}^n P_i=P_0\sum_{i=0}^n(1+r)^i=P_0\frac{1-(1+r)^{n+1}}{1-(1+r)}=P_0\frac{(1+r)^{n+1}-1}{r}.\]
This sum is exactly what is needed whenever the total accumulated quantity across all steps matters, not just its value at the end — for example, total revenue collected over \(n\) years of \(r\) percent growth per year, rather than only the final year's revenue.
Euler Product
\[\sum\limits_{i=1}^\infty \frac{1}{i^x} = \prod\limits_p\frac{1}{1 - p^{-x}}\]
The Euler Product relates natural numbers to prime numbers. The equation is true for any \(x>1\). On the left hand side the sum is also called the Riemann zeta function:
\[\zeta(x) = \sum\limits_{i=1}^\infty \frac{1}{i^x}\]
Harmonic series
\[\sum\limits_{i=1}^\infty\frac{1}{i}=\infty\]
The Harmonic series diverges, because new packages that are \(\geq\frac{1}{2}\) can always be put together:
\[\begin{array}{rl} &= 1+\underbrace{\frac{1}{2}+\frac{1}{3}}_{>\frac{1}{2}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}_{>\frac{1}{2}}+...+\frac{1}{n}\\&<1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...\end{array}\]
The Harmonic series is also \(\zeta(1) = \sum\limits_{i=1}^\infty\frac{1}{i^1} = \infty\), which isn’t defined.
Basler Problem
\[\sum\limits_{i=1}^\infty\frac{1}{i^2}=\frac{\pi^2}{6}\]
Euler proofed that the General Harmonic Series, called Basler Problem converges:
\[1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots = \frac{\pi^2}{6}\]
The Basler Problem is also \(\zeta(2) = \sum\limits_{i=1}^\infty\frac{1}{i^2}\).
General Harmonic Series
\[\sum\limits_{i=1}^\infty\frac{1}{i^x}\]
The General Harmonic Series converges \(\forall x>1\), which means that \(\zeta(x)\) converges \(\forall x>1\).
In the classical real-variable view one often writes \(\zeta(k)=\sum_{i=1}^{\infty}i^{-k}\) with \(k\in\mathbb{R}\). More generally, the same object extends to complex arguments \(k\in\mathbb{C}\) (Riemann zeta function).
The finite general Harmonic Series is defined as
\[H_i^k = \sum\limits_{i=1}^n\frac{1}{i^k}\]
Convergence / Divergence criteria
Divergence criterion
The necessary but not sufficient criterion for convergence of a series \((a_n)_{n\geq k}\) is \(\lim\limits_{n\to\infty}a_n = 0\)
If \((a_n)_{n\geq k}\) is no null sequence, the series \(\sum\limits_{i=k}^\infty a_i\) diverges.
Example: \(\sum\limits_{i=1}^\infty\left(1+\frac{1}{i}\right)\) is divergent, since \(\lim\limits_{i\to\infty}\left(1+\frac{1}{i}\right)=1\)
Weierstrass comparison test (majorant criterion)
Let \((a_n)_{n\geq k}\), \((b_n)_{n\geq k}\) be sequnces with \(|a_n|\leq b_n\), then: If \(\sum\limits_{i=k}^\infty b_i\) is convergent then \(\sum\limits_{i=k}^\infty |a_i|\) is also convergent as well as \(\sum\limits_{i=k}^\infty a_i\).
Leibniz test for alternating series
Let \((a_n)_{n\geq k}\) be a monotonically decreasing null sequence with \(a_i\geq 0\forall i\), then the alternating series \(\sum\limits_{i=k}^\infty (-1)^ia_i\) converges.
Absolute Convergence
\(\sum\limits_{i=k}^\infty a_i\) is called absolute convergent, if \(\sum\limits_{i=k}^\infty|a_i|\) converges. If a series converges absolute \(\Rightarrow\) the series converges. \(\Leftarrow\) does not hold, see alternating Harmonic series.
Root test
If there exists a \(q<1\) and an index \(i_0\) for which \(\sqrt[i]{|a_i|}\leq q\forall i\geq i_0\) holds, the series \(\sum\limits_{i=k}^\infty a_i\) converges absolute.
Please note: \(\sqrt[i]{|a_i|}< 1\) is not enough! For example the Harmonic series \(\sqrt[i]{\frac{1}{i}}\to 1\), but we can’t find a \(q<1\).
If \(\sqrt[i]{|a_i|}\geq 1\) holds for endless \(i\), the series diverges.
d’Alembert’s ratio test (Quotient Criterion)
If there exists a \(q<1\) and an index \(i_0\) for which \(\left|\frac{a_{i+1}}{a_i}\right|\leq q\forall i\geq i_0\) holds, the series \(\sum\limits_{i=k}^\infty a_i\) converges absolute.
Please note: \(\left|\frac{a_{i+1}}{a_i}\right|< 1\) is not enough!
For \(\left|\frac{a_{i+1}}{a_i}\right|\geq 1\) no general statement is possible.