Real series/Introduction/Section

Definition

Let ${}{\left(a_{k}\right)}_{k\in \mathbb {N} }$ be a sequence of real numbers. The series ${}\sum _{k=0}^{\infty }a_{k}$ is the sequence ${}{\left(s_{n}\right)}_{n\in \mathbb {N} }$ of the partial sums

${}s_{n}:=\sum _{k=0}^{n}a_{k}\,.$

If the sequence ${}{\left(s_{n}\right)}_{n\in \mathbb {N} }$ converges, then we say that the series converges. In this case, we write also

$\sum _{k=0}^{\infty }a_{k}$

for its limit,

and this limit is called the sum of the series.

All concepts for sequences carry over to series if we consider a series ${}\sum _{k=0}^{\infty }a_{k}$ as the sequence of its partial sums ${}s_{n}=\sum _{k=0}^{n}a_{k}$. Like for sequences, it might happen that the sequence does not start with ${}k=0$ but later.

Example

We want to compute the series

$\sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}.$

For this, we give a formula for the ${}n$-th partial sum. We have

${}s_{n}=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}{\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)}=1-{\frac {1}{n+1}}={\frac {n}{n+1}}\,.$

This sequence converges to ${}1$, so that the series converges and its sum equals ${}1$.

Lemma

Let

$\sum _{k=0}^{\infty }a_{k}{\text{ and }}\sum _{k=0}^{\infty }b_{k}$

denote convergent series of real numbers with sums ${}s$ and ${}t$

respectively. Then the following statements hold.

1. The series ${}\sum _{k=0}^{\infty }c_{k}$ given by ${}c_{k}:=a_{k}+b_{k}$ is also convergent and its sum is ${}s+t$.
2. For ${}r\in \mathbb {R}$ also the series ${}\sum _{k=0}^{\infty }d_{k}$ given by ${}d_{k}:=ra_{k}$ is convergent and its sum is ${}rs$.

Proof

See exercise.

$\Box$

Lemma

Let

$\sum _{k=0}^{\infty }a_{k}$

be a series of real numbers. Then the series is convergent if and only if the following Cauchy-criterion holds: For every ${}\epsilon >0$ there exists some ${}n_{0}$ such that for all

${}n\geq m\geq n_{0}\,$

the estimate

${}\vert {\sum _{k=m}^{n}a_{k}}\vert \leq \epsilon \,$

holds.

Proof

See exercise.

$\Box$

Lemma

Let

$\sum _{k=0}^{\infty }a_{k}$

denote a convergent series of real numbers. Then

${}\lim _{k\rightarrow \infty }a_{k}=0\,.$

Proof

This follows directly from fact.

$\Box$

Nikolaus of Oresme (1330-1382) proved that the harmonic series diverges.

It is therefore a necessary condition for the convergence of a series that its members form a null sequence. This condition is not sufficient, as the harmonic series shows.

Example

The harmonic series is the series

$\sum _{k=1}^{\infty }{\frac {1}{k}}.$

So this series is about the "infinite sum“ of the unit fractions

$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+\ldots .$

This series diverges: For the ${}2^{n}$ numbers ${}k=2^{n}+1,\ldots ,2^{n+1}$, we have

${}\sum _{k=2^{n}+1}^{2^{n+1}}{\frac {1}{k}}\geq \sum _{k=2^{n}+1}^{2^{n+1}}{\frac {1}{2^{n+1}}}=2^{n}{\frac {1}{2^{n+1}}}={\frac {1}{2}}\,.$

Therefore,

${}\sum _{k=1}^{2^{n+1}}{\frac {1}{k}}=1+\sum _{i=0}^{n}\left(\sum _{k=2^{i}+1}^{2^{i+1}}{\frac {1}{k}}\right)\geq 1+(n+1){\frac {1}{2}}\,.$

Hence, the sequence of the partial sums is unbounded, and so, due to fact, not convergent.

The divergence of the harmonic series implies that one can construct with equal building bricks an arbitrary large overhang.

The following statement is called Leibniz criterion for alternating series.

Theorem

Let ${}{\left(x_{k}\right)}_{k\in \mathbb {N} }$ be an decreasing null sequence of nonnegative real numbers. Then the series ${}\sum _{k=0}^{\infty }(-1)^{k}x_{k}$

converges.

Proof

This proof was not presented in the lecture.

$\Box$