Silverman–Toeplitz theorem: Difference between revisions

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=== Proving 2. ===

=== Proving 2. ===

<math>

<math>

\lim_{n \to \infty} \left( z_{n} – z_{\infty} \right) = 0

\lim_{n \to \infty} \left( z_{} – z_{\infty} \right) = 0

</math>. Applying the already proven statement yields <math>\lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \left( z_{n} – z_{\infty} \right) \big) = 0

</math>. Applying the already proven statement yields <math>\lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \left( z_{} – z_{\infty} \right) \big) = 0

</math>. Finally,

</math>

<math>\lim_{n \to \infty} S_{n}

<math>\lim_{n \to \infty} S_{n}

= \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} z_{n} \big)

= \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} z_{} \big)

= \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \left( z_{n} – z_{\infty} \right) \big) + z_{\infty} \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \big)

= \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \left( z_{} – z_{\infty} \right) \big) + z_{\infty} \lim_{n \to \infty} \sum_{m=1}^{n} \big( a_{n,m} \big)

= 0 + z_{\infty} \cdot 1 = z_{\infty}

= 0 + z_{\infty} \cdot 1 = z_{\infty}

</math>, which completes the proof.

</math>, which completes the proof.

Theorem of summability methods

In mathematics, the Silverman–Toeplitz theorem, first proved by Otto Toeplitz, is a result in series summability theory characterizing matrix summability methods that are regular. A regular matrix summability method is a linear sequence transformation that preserves the limits of convergent sequences.^[1] The linear sequence transformation can be applied to the divergent sequences of partial sums of divergent series to give those series generalized sums.

An infinite matrix ${\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}$ with complex-valued entries defines a regular matrix summability method if and only if it satisfies all of the following properties:

${\displaystyle {\begin{aligned}&\lim _{i\to \infty }a_{i,j}=0\quad j\in \mathbb {N} &&{\text{(Every column sequence converges to 0.)}}\\[3pt]&\lim _{i\to \infty }\sum _{j=0}^{\infty }a_{i,j}=1&&{\text{(The row sums converge to 1.)}}\\[3pt]&\sup _{i}\sum _{j=0}^{\infty }\vert a_{i,j}\vert <\infty &&{\text{(The absolute row sums are bounded.)}}\end{aligned}}}$

An example is Cesàro summation, a matrix summability method with

${\displaystyle a_{mn}={\begin{cases}{\frac {1}{m}}&n\leq m\\0&n>m\end{cases}}={\begin{pmatrix}1&0&0&0&0&\cdots \\{\frac {1}{2}}&{\frac {1}{2}}&0&0&0&\cdots \\{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&0&0&\cdots \\{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&0&\cdots \\{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&\cdots \\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \\\end{pmatrix}}.}$

Formal statement

Let the aforementioned inifinite matrix ${\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}$ of complex elements satisfy the following conditions:

1. ${\displaystyle \lim _{i\to \infty }a_{i,j}=0}$ for every fixed ${\displaystyle j\in \mathbb {N} }$.
2. ${\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty }$;

and ${\displaystyle z_{n}}$ be a sequence of complex numbers that converges to ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }}$. We denote ${\displaystyle S_{n}}$ as the weighted sum sequence: ${\displaystyle S_{n}=\sum _{m=1}^{n}{\left(a_{n,m}z_{m}\right)}}$.

Then the following results hold:

1. If ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0}$, then ${\displaystyle \lim _{n\to \infty }{S_{n}}=0}$.
2. If ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0}$ and ${\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1}$, then ${\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }}$.^[2]

Proof

Proving 1.

For the fixed ${\displaystyle j\in \mathbb {N} }$ the complex sequences ${\displaystyle z_{n}}$, ${\displaystyle S_{n}}$ and ${\displaystyle a_{i,j}}$ approach zero if and only if the real-values sequences ${\displaystyle \left|z_{n}\right|}$, ${\displaystyle \left|S_{n}\right|}$ and ${\displaystyle \left|a_{i,j}\right|}$ approach zero respectively. We also introduce ${\displaystyle M=1+\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0}$.

Since ${\displaystyle \left|z_{n}\right|\to 0}$, for prematurely chosen ${\displaystyle \varepsilon >0}$ there exists ${\displaystyle N_{\varepsilon }\in \mathbb {N} }$, so for every ${\displaystyle n>N_{\varepsilon }}$ we have ${\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}}$. Next, for some ${\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }}$ it’s true, that ${\displaystyle \sum _{m=1}^{n}|a_{n,m}|<{\frac {\varepsilon }{2\left(\max _{m\leq N_{\varepsilon }}|z_{m}|+1\right)}}}$ for every ${\displaystyle n>N_{a}\left(\varepsilon \right)}$. Therefore, for every ${\displaystyle n>N_{a}\left(\varepsilon \right)}$

${\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{m}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)+\sum _{m=N_{\varepsilon }+1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)<\\&<\max _{1\leq m\leq N_{\varepsilon }}(|z_{m}|)\cdot \sum _{m=1}^{N_{\varepsilon }}|a_{n,m}|+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }+1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\sum _{m=1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\cdot M=\varepsilon \end{aligned}}}$

which means, that both sequences ${\displaystyle \left|S_{n}\right|}$ and ${\displaystyle S_{n}}$ converge zero.^[3]

Proving 2.

${\displaystyle \lim _{n\to \infty }\left(z_{m}-z_{\infty }\right)=0}$. Applying the already proven statement yields . Finally,${\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}=0}$

${\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}z_{m}{\big )}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}+z_{\infty }\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}{\big )}=0+z_{\infty }\cdot 1=z_{\infty }}$, which completes the proof.

References

Citations

Further reading

• Toeplitz, Otto (1911) “Über allgemeine lineare Mittelbildungen.” Prace mat.-fiz., 22, 113–118 (the original paper in German)
• Silverman, Louis Lazarus (1913) “On the definition of the sum of a divergent series.” University of Missouri Studies, Math. Series I, 1–96
• Hardy, G. H. (1949), Divergent Series, Oxford: Clarendon Press, 43-48.
• Boos, Johann (2000). Classical and modern methods in summability. New York: Oxford University Press. ISBN 019850165X.