Silverman–Toeplitz theorem: Difference between revisions

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=== Proving 2. ===

<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}

= \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}

</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 $( a i , j ) i , j ∈ N {\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/6d88c66209663e5ba74a1454f12e90103914ae2b" aria-hidden="true" alt="{\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}" width="3957.2" height="1295.7">$ 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}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e27963b6a547742832d39a7e1f52ae3a02181586" aria-hidden="true" alt="{\displaystyle {\begin{aligned}&amp;\lim _{i\to \infty }a_{i,j}=0\quad j\in \mathbb {N} &amp;&{\text{(Every column sequence converges to 0.)}}\\[3pt]&amp;\lim _{i\to \infty }\sum _{j=0}^{\infty }a_{i,j}=1&amp;&{\text{(The row sums converge to 1.)}}\\[3pt]&amp;\sup _{i}\sum _{j=0}^{\infty }\vert a_{i,j}\vert &lt;\infty &amp;&{\text{(The absolute row sums are bounded.)}}\end{aligned}}}" width="28335.1" height="8758.6">

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}}.} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/1ae064065b0e4d6f72e229a0155b4b91693d3e48" aria-hidden="true" alt="{\displaystyle a_{mn}={\begin{cases}{\frac {1}{m}}&amp;n\leq m\\0&amp;n&gt;m\end{cases}}={\begin{pmatrix}1&amp;0&amp;0&amp;0&amp;0&amp;\cdots \\{\frac {1}{2}}&{\frac {1}{2}}&amp;0&amp;0&amp;0&amp;\cdots \\{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&amp;0&amp;0&amp;\cdots \\{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&amp;0&amp;\cdots \\{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&amp;\cdots \\\vdots &amp;\vdots &amp;\vdots &amp;\vdots &amp;\vdots &amp;\ddots \\\end{pmatrix}}.}" width="22416.8" height="10265.5">

Formal statement

Let the aforementioned inifinite matrix $( a i , j ) i , j ∈ N {\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/6d88c66209663e5ba74a1454f12e90103914ae2b" aria-hidden="true" alt="{\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}" width="3957.2" height="1295.7">$ of complex elements satisfy the following conditions:

$lim i → ∞ a i , j = 0 {\displaystyle \lim _{i\to \infty }a_{i,j}=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a42a7d8f9871c213775354bcf2826ea2cc180816" aria-hidden="true" alt="{\displaystyle \lim _{i\to \infty }a_{i,j}=0}" width="5022.9" height="1654.5">$ for every fixed $j ∈ N {\displaystyle j\in \mathbb {N} } <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b54b0f8d5a5eb196ce103460f69c9eb3ec9393fb" aria-hidden="true" alt="{\displaystyle j\in \mathbb {N} }" width="2369.6" height="1080.4">$ .
$sup i ∈ N ∑ j = 1 i | a i , j | < ∞ {\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty } <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/bb568aab0bcb467ab829b8830c49f1198321d43d" aria-hidden="true" alt="{\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty }" width="7539.3" height="3233.2">$ ;

and $z n {\displaystyle z_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5e8228e40cfaa6cb1f163c066ae7054faeff8c7a" aria-hidden="true" alt="{\displaystyle z_{n}}" width="990.1" height="865.1">$ be a sequence of complex numbers that converges to $lim n → ∞ z n = z ∞ {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c129dcee7056573f08e369ec2a7964ae9690c18a" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }}" width="5603.3" height="1582.7">$ . We denote $S n {\displaystyle S_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9f049ac28d4ac8097b625f9d71c1f22b2ebd1bc4" aria-hidden="true" alt="{\displaystyle S_{n}}" width="1138.1" height="1080.4">$ as the weighted sum sequence: $S n = ∑ m = 1 n ( a n , m z m ) {\displaystyle S_{n}=\sum _{m=1}^{n}{\left(a_{n,m}z_{m}\right)}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8751bea444aaf5947298912660ccffa4d2bf5d59" aria-hidden="true" alt="{\displaystyle S_{n}=\sum _{m=1}^{n}{\left(a_{n,m}z_{m}\right)}}" width="8002.4" height="2946.1">$ .

Then the following results hold:

If $lim n → ∞ z n = z ∞ = 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/097030f8bee1f603e363e76c021665480ea3013d" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0}" width="7437.9" height="1582.7">$ , then $lim n → ∞ S n = 0 {\displaystyle \lim _{n\to \infty }{S_{n}}=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/0a985a1243efb35bd9d4f5471bf0aa0a737f9970" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }{S_{n}}=0}" width="4978.9" height="1582.7">$ .
If $lim n → ∞ z n = z ∞ ≠ 0 {\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d9266fb3f947637c7af168c21fa0f11bed9c8a63" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0}" width="7437.9" height="1582.7">$ and $lim i → ∞ ∑ j = 1 i a i , j = 1 {\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ecd5b6dd6efd8a52ed96fd62390c723306f7b33e" aria-hidden="true" alt="{\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1}" width="6634" height="3233.2">$ , then $lim n → ∞ S n = z ∞ {\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/43a42f4b16f06785d98b062693bb9bedeae9adf1" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }}" width="5751.3" height="1582.7">$ .^[2]

Proof

Proving 1.

For the fixed $j ∈ N {\displaystyle j\in \mathbb {N} } <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/b54b0f8d5a5eb196ce103460f69c9eb3ec9393fb" aria-hidden="true" alt="{\displaystyle j\in \mathbb {N} }" width="2369.6" height="1080.4">$ the complex sequences $z n {\displaystyle z_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5e8228e40cfaa6cb1f163c066ae7054faeff8c7a" aria-hidden="true" alt="{\displaystyle z_{n}}" width="990.1" height="865.1">$ , $S n {\displaystyle S_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9f049ac28d4ac8097b625f9d71c1f22b2ebd1bc4" aria-hidden="true" alt="{\displaystyle S_{n}}" width="1138.1" height="1080.4">$ and $a i , j {\displaystyle a_{i,j}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/4bb5a346f58c6568306a02596dd318d1b7e6b2c2" aria-hidden="true" alt="{\displaystyle a_{i,j}}" width="1362.4" height="1008.6">$ approach zero if and only if the real-values sequences $| z n | {\displaystyle \left|z_{n}\right|} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/8f5545535a1cb6c0724621a49c13e65585d360b1" aria-hidden="true" alt="{\displaystyle \left|z_{n}\right|}" width="1547.1" height="1223.9">$ , $| S n | {\displaystyle \left|S_{n}\right|} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/50e72d83733eec2a26188d795767eaefaede286b" aria-hidden="true" alt="{\displaystyle \left|S_{n}\right|}" width="1695.1" height="1223.9">$ and $| a i , j | {\displaystyle \left|a_{i,j}\right|} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/5b68ef912102677ac61847e5af24953d01a58cc5" aria-hidden="true" alt="{\displaystyle \left|a_{i,j}\right|}" width="1919.4" height="1295.7">$ approach zero respectively. We also introduce $M = 1 + sup i ∈ N ∑ j = 1 i | a i , j | > 0 {\displaystyle M=1+\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/97c00b096630d9f6b3a89b7aa2b97f7769dcdaa3" aria-hidden="true" alt="{\displaystyle M=1+\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0}" width="11148.3" height="3233.2">$ .

Since $| z n | → 0 {\displaystyle \left|z_{n}\right|\to 0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/26330b1195266f293841d933de09f8f458dd42ab" aria-hidden="true" alt="{\displaystyle \left|z_{n}\right|\to 0}" width="3603.7" height="1223.9">$ , for prematurely chosen $ε > 0 {\displaystyle \varepsilon >0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/e04ec3670b50384a3ce48aca42e7cc5131a06b12" aria-hidden="true" alt="{\displaystyle \varepsilon >0}" width="2301.1" height="936.9">$ there exists $N ε ∈ N {\displaystyle N_{\varepsilon }\in \mathbb {N} } <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2ce9271e20f73756eae29d5bd966446ba039cb8c" aria-hidden="true" alt="{\displaystyle N_{\varepsilon }\in \mathbb {N} }" width="3178.9" height="1080.4">$ , so for every $n > N ε {\displaystyle n>N_{\varepsilon }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/6c726dac045d00dd1af2ca514a0a5d638133594c" aria-hidden="true" alt="{\displaystyle n>N_{\varepsilon }}" width="3167.9" height="1080.4">$ we have $| z n | < ε 2 M {\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a4a4837019d63abc43debed1bafc690e072428a3" aria-hidden="true" alt="{\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}}" width="4793.2" height="2013.3">$ . Next, for some $N a = N a ( ε ) > N ε {\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f967a93d868f16fa070d1368feb8755ed3eba61e" aria-hidden="true" alt="{\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }}" width="7869.5" height="1223.9">$ it’s true, that $∑ m = 1 n | a n , m | < ε 2 ( max m ≤ N ε | z m | + 1 ) {\displaystyle \sum _{m=1}^{n}|a_{n,m}|<{\frac {\varepsilon }{2\left(\max _{m\leq N_{\varepsilon }}|z_{m}|+1\right)}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/66cb644247238386c41795f000aaf965465919c6" aria-hidden="true" alt="{\displaystyle \sum _{m=1}^{n}|a_{n,m}|<{\frac {\varepsilon }{2\left(\max _{m\leq N_{\varepsilon }}|z_{m}|+1\right)}}}" width="14936.4" height="2946.1">$ for every $n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cbae5323e5085b329bc67236c4a75f065e312a6e" aria-hidden="true" alt="{\displaystyle n>N_{a}\left(\varepsilon \right)}" width="4624.6" height="1223.9">$ . Therefore, for every $n > N a ( ε ) {\displaystyle n>N_{a}\left(\varepsilon \right)} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/cbae5323e5085b329bc67236c4a75f065e312a6e" aria-hidden="true" alt="{\displaystyle n>N_{a}\left(\varepsilon \right)}" width="4624.6" height="1223.9">$

$| S n | = | ∑ m = 1 n ( a n , m z m ) | ⩽ ∑ m = 1 n ( | a n , m | ⋅ | z m | ) = ∑ m = 1 N ε ( | a n , m | ⋅ | z m | ) + ∑ m = N ε + 1 n ( | a n , m | ⋅ | z m | ) < < max 1 ≤ m ≤ N ε ( | z m | ) ⋅ ∑ m = 1 N ε | a n , m | + ε 2 M ∑ m = N ε + 1 n | a n , m | ⩽ ε 2 + ε 2 M ∑ m = 1 n | a n , m | ⩽ ε 2 + ε 2 M ⋅ M = ε {\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{m}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)+\sum _{m=N_{\varepsilon }+1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)<\\&<\max _{1\leq m\leq N_{\varepsilon }}(|z_{m}|)\cdot \sum _{m=1}^{N_{\varepsilon }}|a_{n,m}|+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }+1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\sum _{m=1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\cdot M=\varepsilon \end{aligned}}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d5ec65b706cf03d5083c4d8c7342bfb1f7b627dc" aria-hidden="true" alt="{\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{m}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)+\sum _{m=N_{\varepsilon }+1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{m}\right|\right)<\\&<\max _{1\leq m\leq N_{\varepsilon }}(|z_{m}|)\cdot \sum _{m=1}^{N_{\varepsilon }}|a_{n,m}|+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }+1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\sum _{m=1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\cdot M=\varepsilon \end{aligned}}}" width="39409.9" height="6677.6">$

which means, that both sequences $| S n | {\displaystyle \left|S_{n}\right|} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/50e72d83733eec2a26188d795767eaefaede286b" aria-hidden="true" alt="{\displaystyle \left|S_{n}\right|}" width="1695.1" height="1223.9">$ and $S n {\displaystyle S_{n}} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9f049ac28d4ac8097b625f9d71c1f22b2ebd1bc4" aria-hidden="true" alt="{\displaystyle S_{n}}" width="1138.1" height="1080.4">$ converge zero.^[3]

Proving 2.

$lim n → ∞ ( z m − z ∞ ) = 0 {\displaystyle \lim _{n\to \infty }\left(z_{m}-z_{\infty }\right)=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/257b097296800f1f85e4e23f599da3f49e610be5" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }\left(z_{m}-z_{\infty }\right)=0}" width="8302.4" height="1654.5">$ . Applying the already proven statement yields . Finally, $lim n → ∞ ∑ m = 1 n ( a n , m ( z m − z ∞ ) ) = 0 {\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}=0} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/f219561659301c2c7e44382e6bf7a5509f58158f" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}=0}" width="13117.2" height="2946.1">$

$lim n → ∞ S n = lim n → ∞ ∑ m = 1 n ( a n , m z m ) = lim n → ∞ ∑ m = 1 n ( a n , m ( z m − z ∞ ) ) + z ∞ lim n → ∞ ∑ m = 1 n ( a n , m ) = 0 + z ∞ ⋅ 1 = z ∞ {\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}z_{m}{\big )}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}+z_{\infty }\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}{\big )}=0+z_{\infty }\cdot 1=z_{\infty }} <img decoding="async" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ec629ed9eab54e57b26c0a70c4b2c1da52d23b88" aria-hidden="true" alt="{\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}z_{m}{\big )}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{m}-z_{\infty }\right){\big )}+z_{\infty }\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}{\big )}=0+z_{\infty }\cdot 1=z_{\infty }}" width="42080.6" height="2946.1">$ , which completes the proof.

References

Citations

^ Silverman–Toeplitz theorem, by Ruder, Brian, Published 1966, Call number LD2668 .R4 1966 R915, Publisher Kansas State University, Internet Archive
^ Linero, Antonio; Rosalsky, Andrew (2013-07-01). “On the Toeplitz Lemma, Convergence in Probability, and Mean Convergence” (PDF). Stochastic Analysis and Applications. 31 (4): 684–694. doi:10.1080/07362994.2013.799406. ISSN 0736-2994. Retrieved 2024-11-17.
^ Ljashko, Ivan Ivanovich; Bojarchuk, Alexey Klimetjevich; Gaj, Jakov Gavrilovich; Golovach, Grigory Petrovich (2001). Математический анализ: введение в анализ, производная, интеграл. Справочное пособие по высшей математике [Mathematical analysis: the introduction into analysis, derivatives, integrals. The handbook to mathematical analysis.] (in Russian). Vol. 1 (1st ed.). Moskva: Editorial URSS. p. 58. ISBN 978-5-354-00018-0.

Silverman–Toeplitz theorem: Difference between revisions

Formal statement