Cauchy's limit theorem

The Cauchy limit theorem was first discovered by the French mathematician Augustin-Louis Cauchy formulated. It is a special case of the more general theorem of Cesàro – Stolz and says: From the convergence of a sequence of numbers follows the convergence of the Cesàro means of the sequence towards the same limit value. Or: it follows . ${\ displaystyle a_ {n} \ to a}$ ${\ displaystyle (a_ {1} + \ cdots + a_ {n}) / n \ \ to a}$

Related results and extensions

If one considers a weighted mean instead of the usual arithmetic mean , then the convergence of the original sequence also implies the convergence of the weighted means, that is, the following theorem applies:

Be an arbitrary sequence with and a sequence of positive numbers , then the following applies: . ${\ displaystyle (a_ {n})}$ ${\ displaystyle a_ {n} \ rightarrow a}$ ${\ displaystyle (p_ {n})}$ ${\ displaystyle {\ frac {1} {p_ {1} + \ ldots + p_ {n}}} \ rightarrow 0}$ ${\ displaystyle {\ frac {p_ {1} a_ {1} + \ ldots + p_ {n} a_ {n}} {p_ {1} + \ ldots + p_ {n}}} \ rightarrow a}$

An analogous theorem also applies to the geometric mean :

Be a sequence with , then the following applies: . ${\ displaystyle (a_ {n})}$ ${\ displaystyle a_ {n} \ rightarrow a}$ ${\ displaystyle {\ sqrt [{n}] {a_ {1} \ cdot a_ {2} \ cdot \ ldots \ cdot a_ {n}}} \ \ rightarrow a}$

Proof of Cauchy's limit theorem

Be arbitrary and chosen so that applies to everyone . Because there is a with for . ${\ displaystyle \ varepsilon> 0}$ ${\ displaystyle N \ in \ mathbb {N}}$ ${\ displaystyle | a_ {k} -a | <{\ tfrac {\ varepsilon} {2}}}$ ${\ displaystyle k \ geq N}$
${\ displaystyle \ lim _ {n \ to \ infty} {\ frac {1} {n}} \ sum _ {k = 1} ^ {N} (a_ {k} -a) = 0}$ ${\ displaystyle M \ in \ mathbb {N}}$ ${\ displaystyle \ left | {\ frac {1} {n}} \ sum _ {k = 1} ^ {N} (a_ {k} -a) \ right | <{\ frac {\ varepsilon} {2} }}$ ${\ displaystyle n \ geq M}$