General Cesaro summation with weightCan you please check my Cesaro means proofDoes $sumlimits_n=1^inftyfrac1n^2(a_1+cdots+ a_n)$ converges when $a_n$ converges?limit of a geometric meanShow that $lim_ntoinftysqrt[sum_limitsk=0^nlambda_k]prod_limitsk=0^na_k^lambda_k = ell =lim_ntoinfty a_n$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsGiven a convergent series $sumlimits_n=1^infty a_k$, show that $limlimits_ntoinftyfrac1nsumlimits_k=1^n a_k = 0$Prove that if $s_n$ is bounded and monotonic, then $t_n =(s_1 + cdots+ s_n)/n$ converges to the same limit as $s_n$Show that $displaystyle lim_n to inftysum_i=1^n frac((n+1)-i)a_in^2 = fraca2 $Show $limlimits_n to infty frac1n^2 sumlimits_k=1^n frack+1log(k+1) = 0$Existence of $p_n$ such that $lim_n to infty fraca_1+a_2+dots+a_p_n+1a_1+a_2+dots+a_p_n=l$A generalized Cesaro Mean.. any ideas?prove that if $lim limits_n to inftyF( a_n)=ell$, then $lim limits_x to inftyF( x)=ell$Cesaro summationUnderstanding Cesaro summation proofProve $a_n$ converges to zero with $a_n+1leqslant (1-lambda_n)a_n+b_n+c_n$Prove that $limlimits_ntoinftyleft(sum_k=0^nlambda_kright)left(sum_k=0^nfraclambda_k a_kright)^-1= lim_ntoinfty a_n$Show that $lim_ntoinftysqrt[sum_limitsk=0^nlambda_k]prod_limitsk=0^na_k^lambda_k = ell =lim_ntoinfty a_n$If $prod _n=1^infty a_nneq 0$ converges then $limlimits_ntoinfty frac1sum_limitsk=0^nlambda_ksum_k=0^nlambda_k a_k =1$Prove that $prod_limitsn = k_0^infty (1 - a_n) ;$ converges to positive valueCesaro Means decrease slower than the sequence
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General Cesaro summation with weight
Can you please check my Cesaro means proofDoes $sumlimits_n=1^inftyfrac1n^2(a_1+cdots+ a_n)$ converges when $a_n$ converges?limit of a geometric meanShow that $lim_ntoinftysqrt[sum_limitsk=0^nlambda_k]prod_limitsk=0^na_k^lambda_k = ell =lim_ntoinfty a_n$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsGiven a convergent series $sumlimits_n=1^infty a_k$, show that $limlimits_ntoinftyfrac1nsumlimits_k=1^n a_k = 0$Prove that if $s_n$ is bounded and monotonic, then $t_n =(s_1 + cdots+ s_n)/n$ converges to the same limit as $s_n$Show that $displaystyle lim_n to inftysum_i=1^n frac((n+1)-i)a_in^2 = fraca2 $Show $limlimits_n to infty frac1n^2 sumlimits_k=1^n frack+1log(k+1) = 0$Existence of $p_n$ such that $lim_n to infty fraca_1+a_2+dots+a_p_n+1a_1+a_2+dots+a_p_n=l$A generalized Cesaro Mean.. any ideas?prove that if $lim limits_n to inftyF( a_n)=ell$, then $lim limits_x to inftyF( x)=ell$Cesaro summationUnderstanding Cesaro summation proofProve $a_n$ converges to zero with $a_n+1leqslant (1-lambda_n)a_n+b_n+c_n$Prove that $limlimits_ntoinftyleft(sum_k=0^nlambda_kright)left(sum_k=0^nfraclambda_k a_kright)^-1= lim_ntoinfty a_n$Show that $lim_ntoinftysqrt[sum_limitsk=0^nlambda_k]prod_limitsk=0^na_k^lambda_k = ell =lim_ntoinfty a_n$If $prod _n=1^infty a_nneq 0$ converges then $limlimits_ntoinfty frac1sum_limitsk=0^nlambda_ksum_k=0^nlambda_k a_k =1$Prove that $prod_limitsn = k_0^infty (1 - a_n) ;$ converges to positive valueCesaro Means decrease slower than the sequence
$begingroup$
Assume that $a_nto ell $ is a convergent sequence of complex numbers and $lambda_n$ is a sequence of positive real numbers such that $sumlimits_k=0^inftylambda_k = infty$
Then, show that,
$$lim_ntoinfty frac1sum_limitsk=0^nlambda_k sum_limitsk=0^nlambda_k a_k=ell =lim_ntoinfty a_n$$
(Note that : This is more general than the special case where, $lambda_n= 1$)
calculus real-analysis sequences-and-series limits cesaro-summable
edited Mar 13 at 8:35
BijanDatta
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
add a comment |
$begingroup$
Assume that $a_nto ell $ is a convergent sequence of complex numbers and $lambda_n$ is a sequence of positive real numbers such that $sumlimits_k=0^inftylambda_k = infty$
Then, show that,
$$lim_ntoinfty frac1sum_limitsk=0^nlambda_k sum_limitsk=0^nlambda_k a_k=ell =lim_ntoinfty a_n$$
(Note that : This is more general than the special case where, $lambda_n= 1$)
calculus real-analysis sequences-and-series limits cesaro-summable
edited Mar 13 at 8:35
BijanDatta
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
$begingroup$
Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
$endgroup$
– Harry
Sep 22 '17 at 12:22
$begingroup$
there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
$endgroup$
– Guy Fsone
Sep 22 '17 at 12:24
add a comment |
$begingroup$
Assume that $a_nto ell $ is a convergent sequence of complex numbers and $lambda_n$ is a sequence of positive real numbers such that $sumlimits_k=0^inftylambda_k = infty$
Then, show that,
$$lim_ntoinfty frac1sum_limitsk=0^nlambda_k sum_limitsk=0^nlambda_k a_k=ell =lim_ntoinfty a_n$$
(Note that : This is more general than the special case where, $lambda_n= 1$)
calculus real-analysis sequences-and-series limits cesaro-summable
edited Mar 13 at 8:35
BijanDatta
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
Assume that $a_nto ell $ is a convergent sequence of complex numbers and $lambda_n$ is a sequence of positive real numbers such that $sumlimits_k=0^inftylambda_k = infty$
Then, show that,
$$lim_ntoinfty frac1sum_limitsk=0^nlambda_k sum_limitsk=0^nlambda_k a_k=ell =lim_ntoinfty a_n$$
(Note that : This is more general than the special case where, $lambda_n= 1$)
calculus real-analysis sequences-and-series limits cesaro-summable
calculus real-analysis sequences-and-series limits cesaro-summable
edited Mar 13 at 8:35
BijanDatta
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
edited Mar 13 at 8:35
BijanDatta
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
edited Mar 13 at 8:35
BijanDatta
309113
edited Mar 13 at 8:35
BijanDatta
309113
edited Mar 13 at 8:35
BijanDatta
309113
309113
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
asked Sep 22 '17 at 12:14
Guy FsoneGuy Fsone
17.3k43074
17.3k43074
$begingroup$
Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
$endgroup$
– Harry
Sep 22 '17 at 12:22
$begingroup$
there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
$endgroup$
– Guy Fsone
Sep 22 '17 at 12:24
add a comment |
$begingroup$
Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
$endgroup$
– Harry
Sep 22 '17 at 12:22
$begingroup$
there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
$endgroup$
– Guy Fsone
Sep 22 '17 at 12:24
$begingroup$
Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
$endgroup$
– Harry
Sep 22 '17 at 12:22
$begingroup$
Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
$endgroup$
– Harry
Sep 22 '17 at 12:22
$begingroup$
there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
$endgroup$
– Guy Fsone
Sep 22 '17 at 12:24
$begingroup$
there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
$endgroup$
– Guy Fsone
Sep 22 '17 at 12:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First you can assume that the $(a_n)$ are real, by considering
the sequences
$(operatornameRe a_n)_n$ and $(operatornameIm a_n)_n$
separately.
And then it is an immediate application of the Stolz–Cesàro theorem to
$$
A_n := sum_k=0^n lambda_n a_n quad, quad B_n := sum_k=0^n lambda_n
$$
since $(B_n)$ is strictly increasing and unbounded, and
$$
fracA_n+1 - A_nB_n+1 - B_n = a_n+1 to l
$$
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
$endgroup$
add a comment |
$begingroup$
Let $varepsilon >0$ and $N$such that $|a_k-l|le varepsilon $ for all $k>N$
Then, for $n>N$ we have,
beginsplitleft| fracsum_limitsk=0^nlambda_k a_ksum_limitsk=0^nlambda_k -lright|
&= &left| fracsum_limitsk=0^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&= &left| fracsum_limitsk=0^Nlambda_k (a_k - l)+sum_limitsk=N^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&le & fracMsum_limitsk=0^nlambda_k + fracsum_limitsk=N^nlambda_k underbrace a_k - lright_levarepsilonsum_limitsk=0^nlambda_k \
&le&
fracMsum_limitsk=0^nlambda_k + varepsilonto 0
endsplit
since $sum_limitsk=0^Nlambda_kto infty$.
Where $M= left|sum_limitsk=0^Nlambda_k( a_k-l)right|$
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First you can assume that the $(a_n)$ are real, by considering
the sequences
$(operatornameRe a_n)_n$ and $(operatornameIm a_n)_n$
separately.
And then it is an immediate application of the Stolz–Cesàro theorem to
$$
A_n := sum_k=0^n lambda_n a_n quad, quad B_n := sum_k=0^n lambda_n
$$
since $(B_n)$ is strictly increasing and unbounded, and
$$
fracA_n+1 - A_nB_n+1 - B_n = a_n+1 to l
$$
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
$endgroup$
add a comment |
$begingroup$
First you can assume that the $(a_n)$ are real, by considering
the sequences
$(operatornameRe a_n)_n$ and $(operatornameIm a_n)_n$
separately.
And then it is an immediate application of the Stolz–Cesàro theorem to
$$
A_n := sum_k=0^n lambda_n a_n quad, quad B_n := sum_k=0^n lambda_n
$$
since $(B_n)$ is strictly increasing and unbounded, and
$$
fracA_n+1 - A_nB_n+1 - B_n = a_n+1 to l
$$
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
$endgroup$
add a comment |
$begingroup$
First you can assume that the $(a_n)$ are real, by considering
the sequences
$(operatornameRe a_n)_n$ and $(operatornameIm a_n)_n$
separately.
And then it is an immediate application of the Stolz–Cesàro theorem to
$$
A_n := sum_k=0^n lambda_n a_n quad, quad B_n := sum_k=0^n lambda_n
$$
since $(B_n)$ is strictly increasing and unbounded, and
$$
fracA_n+1 - A_nB_n+1 - B_n = a_n+1 to l
$$
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
$endgroup$
First you can assume that the $(a_n)$ are real, by considering
the sequences
$(operatornameRe a_n)_n$ and $(operatornameIm a_n)_n$
separately.
And then it is an immediate application of the Stolz–Cesàro theorem to
$$
A_n := sum_k=0^n lambda_n a_n quad, quad B_n := sum_k=0^n lambda_n
$$
since $(B_n)$ is strictly increasing and unbounded, and
$$
fracA_n+1 - A_nB_n+1 - B_n = a_n+1 to l
$$
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
answered Sep 22 '17 at 12:53
Martin RMartin R
30k33558
30k33558
add a comment |
add a comment |
$begingroup$
Let $varepsilon >0$ and $N$such that $|a_k-l|le varepsilon $ for all $k>N$
Then, for $n>N$ we have,
beginsplitleft| fracsum_limitsk=0^nlambda_k a_ksum_limitsk=0^nlambda_k -lright|
&= &left| fracsum_limitsk=0^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&= &left| fracsum_limitsk=0^Nlambda_k (a_k - l)+sum_limitsk=N^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&le & fracMsum_limitsk=0^nlambda_k + fracsum_limitsk=N^nlambda_k underbrace a_k - lright_levarepsilonsum_limitsk=0^nlambda_k \
&le&
fracMsum_limitsk=0^nlambda_k + varepsilonto 0
endsplit
since $sum_limitsk=0^Nlambda_kto infty$.
Where $M= left|sum_limitsk=0^Nlambda_k( a_k-l)right|$
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
add a comment |
$begingroup$
Let $varepsilon >0$ and $N$such that $|a_k-l|le varepsilon $ for all $k>N$
Then, for $n>N$ we have,
beginsplitleft| fracsum_limitsk=0^nlambda_k a_ksum_limitsk=0^nlambda_k -lright|
&= &left| fracsum_limitsk=0^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&= &left| fracsum_limitsk=0^Nlambda_k (a_k - l)+sum_limitsk=N^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&le & fracMsum_limitsk=0^nlambda_k + fracsum_limitsk=N^nlambda_k underbrace a_k - lright_levarepsilonsum_limitsk=0^nlambda_k \
&le&
fracMsum_limitsk=0^nlambda_k + varepsilonto 0
endsplit
since $sum_limitsk=0^Nlambda_kto infty$.
Where $M= left|sum_limitsk=0^Nlambda_k( a_k-l)right|$
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
add a comment |
$begingroup$
Let $varepsilon >0$ and $N$such that $|a_k-l|le varepsilon $ for all $k>N$
Then, for $n>N$ we have,
beginsplitleft| fracsum_limitsk=0^nlambda_k a_ksum_limitsk=0^nlambda_k -lright|
&= &left| fracsum_limitsk=0^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&= &left| fracsum_limitsk=0^Nlambda_k (a_k - l)+sum_limitsk=N^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&le & fracMsum_limitsk=0^nlambda_k + fracsum_limitsk=N^nlambda_k underbrace a_k - lright_levarepsilonsum_limitsk=0^nlambda_k \
&le&
fracMsum_limitsk=0^nlambda_k + varepsilonto 0
endsplit
since $sum_limitsk=0^Nlambda_kto infty$.
Where $M= left|sum_limitsk=0^Nlambda_k( a_k-l)right|$
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
$endgroup$
Let $varepsilon >0$ and $N$such that $|a_k-l|le varepsilon $ for all $k>N$
Then, for $n>N$ we have,
beginsplitleft| fracsum_limitsk=0^nlambda_k a_ksum_limitsk=0^nlambda_k -lright|
&= &left| fracsum_limitsk=0^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&= &left| fracsum_limitsk=0^Nlambda_k (a_k - l)+sum_limitsk=N^nlambda_k (a_k - l)sum_limitsk=0^nlambda_k right|\
&le & fracMsum_limitsk=0^nlambda_k + fracsum_limitsk=N^nlambda_k underbrace a_k - lright_levarepsilonsum_limitsk=0^nlambda_k \
&le&
fracMsum_limitsk=0^nlambda_k + varepsilonto 0
endsplit
since $sum_limitsk=0^Nlambda_kto infty$.
Where $M= left|sum_limitsk=0^Nlambda_k( a_k-l)right|$
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
answered Sep 22 '17 at 12:58
Guy FsoneGuy Fsone
17.3k43074
17.3k43074
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
add a comment |
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
In the last inequality you mean to say the right-hand side goes to $varepsilon$ as $ntoinfty$.
$endgroup$
– Blackbird
Sep 22 '17 at 13:28
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
$begingroup$
yes of course $varepsilon $is negligible
$endgroup$
– Guy Fsone
Sep 22 '17 at 13:35
add a comment |
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Is there any other restriction? With $lambda_k=k$, the sum diverges (unless perhaps $l=0$).
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– Harry
Sep 22 '17 at 12:22
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there is no problem if the sum diverge. the result is still true see here for instance math.stackexchange.com/questions/2440315/…
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– Guy Fsone
Sep 22 '17 at 12:24