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 . 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 ( a i , j ) i , j ∈ N {\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:
lim i → ∞ a i , j = 0 j ∈ N (Every column sequence converges to 0.) lim i → ∞ ∑ j = 0 ∞ a i , j = 1 (The row sums converge to 1.) sup i ∑ j = 0 ∞ | a i , j | < ∞ (The absolute row sums are bounded.) {\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
a m n = { 1 m n ≤ m 0 n > m = ( 1 0 0 0 0 ⋯ 1 2 1 2 0 0 0 ⋯ 1 3 1 3 1 3 0 0 ⋯ 1 4 1 4 1 4 1 4 0 ⋯ 1 5 1 5 1 5 1 5 1 5 ⋯ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ) . {\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}}.}
Let the aforementioned inifinite matrix ( a i , j ) i , j ∈ N {\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }} of complex elements satisfy the following conditions:
lim i → ∞ a i , j = 0 {\displaystyle \lim _{i\to \infty }a_{i,j}=0} for every fixed j ∈ N {\displaystyle j\in \mathbb {N} } .sup i ∈ N ∑ j = 1 i | a i , j | < ∞ {\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty } ;and z n {\displaystyle z_{n}} be a sequence of complex numbers that converges to lim n → ∞ z n = z ∞ {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }} . We denote S n {\displaystyle S_{n}} as the weighted sum sequence: S n = ∑ m = 1 n ( a n , m z n ) {\displaystyle S_{n}=\sum _{m=1}^{n}{\left(a_{n,m}z_{n}\right)}} .
Then the following results hold:
If lim n → ∞ z n = z ∞ = 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0} , then lim n → ∞ S n = 0 {\displaystyle \lim _{n\to \infty }{S_{n}}=0} . If lim n → ∞ z n = z ∞ ≠ 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0} and lim i → ∞ ∑ j = 1 i a i , j = 1 {\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1} , then lim n → ∞ S n = z ∞ {\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }} .
Proof
Proving 1. For the fixed j ∈ N {\displaystyle j\in \mathbb {N} } the complex sequences z n {\displaystyle z_{n}} , S n {\displaystyle S_{n}} and a i , j {\displaystyle a_{i,j}} approach zero if and only if the real-values sequences | z n | {\displaystyle \left|z_{n}\right|} , | S n | {\displaystyle \left|S_{n}\right|} and | a i , j | {\displaystyle \left|a_{i,j}\right|} approach zero respectively. We also introduce M = sup i ∈ N ∑ j = 1 i | a i , j | > 0 {\displaystyle M=\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0} .
Since | z n | → 0 {\displaystyle \left|z_{n}\right|\to 0} , for prematurely chosen ε > 0 {\displaystyle \varepsilon >0} there exists N ε = N ε ( ε ) {\displaystyle N_{\varepsilon }=N_{\varepsilon }\left(\varepsilon \right)} , so for every n > N ε ( ε ) {\displaystyle n>N_{\varepsilon }\left(\varepsilon \right)} we have | z n | < ε 2 M {\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}} . Next, for some N a = N a ( ε ) > N ε ( ε ) {\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }\left(\varepsilon \right)} it's true, that | a n , m | < M N ε {\displaystyle \left|a_{n,m}\right|<{\frac {M}{N_{\varepsilon }}}} for every n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)} and 1 ⩽ m ⩽ n {\displaystyle 1\leqslant m\leqslant n} . Therefore, for every n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)}
| S n | = | ∑ m = 1 n ( a n , m z n ) | ⩽ ∑ m = 1 n ( | a n , m | ⋅ | z n | ) = ∑ m = 1 N ε ( | a n , m | ⋅ | z n | ) + ∑ m = N ε n ( | a n , m | ⋅ | z n | ) < < N ε ⋅ M N ε ⋅ ε 2 M + ε 2 M ∑ m = N ε n | a n , m | ⩽ ε 2 + ε 2 M ∑ m = 1 n | a n , m | ⩽ ε 2 + ε 2 M ⋅ M = ε {\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{n}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)+\sum _{m=N_{\varepsilon }}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)<\\&<N_{\varepsilon }\cdot {\frac {M}{N_{\varepsilon }}}\cdot {\frac {\varepsilon }{2M}}+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }}^{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 | S n | {\displaystyle \left|S_{n}\right|} and S n {\displaystyle S_{n}} converge zero.
Proving 2. lim n → ∞ ( z n − z ∞ ) = 0 {\displaystyle \lim _{n\to \infty }\left(z_{n}-z_{\infty }\right)=0} . Applying the already proven statement yields lim n → ∞ ∑ m = 1 n ( a n , m ( z n − z ∞ ) ) = 0 {\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{n}-z_{\infty }\right){\big )}=0} . Finally,
lim n → ∞ S n = lim n → ∞ ∑ m = 1 n ( a n , m z n ) = lim n → ∞ ∑ m = 1 n ( a n , m ( z n − z ∞ ) ) + z ∞ lim n → ∞ ∑ m = 1 n ( a n , m ) = 0 + z ∞ ⋅ 1 = z ∞ {\displaystyle \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}\left(z_{n}-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, 1–96Hardy, 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 .