How to prove the Tauber's second theorem?
$begingroup$
The Tauber's second theorem states as follows :
$$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !
calculus
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
$endgroup$
add a comment |
$begingroup$
The Tauber's second theorem states as follows :
$$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !
calculus
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
$endgroup$
add a comment |
$begingroup$
The Tauber's second theorem states as follows :
$$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !
calculus
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
$endgroup$
The Tauber's second theorem states as follows :
$$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !
calculus
calculus
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
asked Dec 15 '18 at 5:41
Alexander LauAlexander Lau
747
747
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think I got one .
$Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.
$Leftarrow$
$$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
letting N tends to infinity , that ends the proof.
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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active
oldest
votes
$begingroup$
I think I got one .
$Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.
$Leftarrow$
$$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
letting N tends to infinity , that ends the proof.
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
$endgroup$
add a comment |
$begingroup$
I think I got one .
$Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.
$Leftarrow$
$$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
letting N tends to infinity , that ends the proof.
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
$endgroup$
add a comment |
$begingroup$
I think I got one .
$Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.
$Leftarrow$
$$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
letting N tends to infinity , that ends the proof.
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
$endgroup$
I think I got one .
$Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.
$Leftarrow$
$$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
letting N tends to infinity , that ends the proof.
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
answered Dec 15 '18 at 6:36
Alexander LauAlexander Lau
747
747
add a comment |
add a comment |
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