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: \(\sum\limits_{i=1}^\infty q^i \begin{cases} |q|<1: \text{ converges to }\frac{1}{1-q} \\ |q|\geq 1: \text{ diverges} \end{cases}\)
• 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 \(\ln 2\)
• \(\sum\limits_{i=1}^\infty\frac{1}{2^i}\) converges to \(1\) since
  

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 \(\forall x>1\).

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.