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Improving the convergence speed in a Leibniz series
Leibniz's alternating series testLeibniz series convergence citeria valid for monotonic increasing zero-sequence?Does Liebniz-Criteria become a necessary condition for convergence if $a_n$ is monotonically decreasing?$sum_n= 0^inftya_n$ converges, what other series must then also converge?Finding the Radius of Convergence of the complex power seriesCounterexample to Leibniz criterion for alternating seriesA counterexample to monotonicity in Leibniz criterion for alternatig seriesConvergence of series of remaindersAsymptotic behavior of series tailHelp with convergence tests for series
$begingroup$
I have read the proof for the Leibniz criterion. Underneath there was a remark that I did not understand it says that if we have a Leibniz series than one can obtain $s$ sometimes faster by calculating the (arithmetic) mean value and with index shift.
With Leibniz series I mean that $(a_n)$ is a monotonically decreasing zero-convergent sequence. And the Leibniz series $sum_n=0^infty(-1)^na_n$ converges. Let $s$ be the value it converges to
My Question why
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
also converges to $s$?
And if
$|s-sum_n=0^k(-1)^na_n|leq a_k+1$
Why is
$|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
Thanks for helping
What I have thought so far is that the if we have
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
then the Right Hand side cannot converge to $s$ it must somehow compensate the influence of the Always constant $a_0/2$ in a way that it converges to $s$. Maybe one can do someting with the distributive law but I am not sure if I am allowed to use it.
In particular if $L=sum (-1)^n frac1n+1$
then also
$L=frac12+frac12big(frac11cdot 2-frac12cdot 3+frac13cdot 4-frac14cdot 5+ …big)$
sequences-and-series convergence
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
$endgroup$
add a comment |
$begingroup$
I have read the proof for the Leibniz criterion. Underneath there was a remark that I did not understand it says that if we have a Leibniz series than one can obtain $s$ sometimes faster by calculating the (arithmetic) mean value and with index shift.
With Leibniz series I mean that $(a_n)$ is a monotonically decreasing zero-convergent sequence. And the Leibniz series $sum_n=0^infty(-1)^na_n$ converges. Let $s$ be the value it converges to
My Question why
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
also converges to $s$?
And if
$|s-sum_n=0^k(-1)^na_n|leq a_k+1$
Why is
$|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
Thanks for helping
What I have thought so far is that the if we have
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
then the Right Hand side cannot converge to $s$ it must somehow compensate the influence of the Always constant $a_0/2$ in a way that it converges to $s$. Maybe one can do someting with the distributive law but I am not sure if I am allowed to use it.
In particular if $L=sum (-1)^n frac1n+1$
then also
$L=frac12+frac12big(frac11cdot 2-frac12cdot 3+frac13cdot 4-frac14cdot 5+ …big)$
sequences-and-series convergence
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
$endgroup$
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34
add a comment |
$begingroup$
I have read the proof for the Leibniz criterion. Underneath there was a remark that I did not understand it says that if we have a Leibniz series than one can obtain $s$ sometimes faster by calculating the (arithmetic) mean value and with index shift.
With Leibniz series I mean that $(a_n)$ is a monotonically decreasing zero-convergent sequence. And the Leibniz series $sum_n=0^infty(-1)^na_n$ converges. Let $s$ be the value it converges to
My Question why
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
also converges to $s$?
And if
$|s-sum_n=0^k(-1)^na_n|leq a_k+1$
Why is
$|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
Thanks for helping
What I have thought so far is that the if we have
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
then the Right Hand side cannot converge to $s$ it must somehow compensate the influence of the Always constant $a_0/2$ in a way that it converges to $s$. Maybe one can do someting with the distributive law but I am not sure if I am allowed to use it.
In particular if $L=sum (-1)^n frac1n+1$
then also
$L=frac12+frac12big(frac11cdot 2-frac12cdot 3+frac13cdot 4-frac14cdot 5+ …big)$
sequences-and-series convergence
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
$endgroup$
I have read the proof for the Leibniz criterion. Underneath there was a remark that I did not understand it says that if we have a Leibniz series than one can obtain $s$ sometimes faster by calculating the (arithmetic) mean value and with index shift.
With Leibniz series I mean that $(a_n)$ is a monotonically decreasing zero-convergent sequence. And the Leibniz series $sum_n=0^infty(-1)^na_n$ converges. Let $s$ be the value it converges to
My Question why
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
also converges to $s$?
And if
$|s-sum_n=0^k(-1)^na_n|leq a_k+1$
Why is
$|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
Thanks for helping
What I have thought so far is that the if we have
$frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)$
then the Right Hand side cannot converge to $s$ it must somehow compensate the influence of the Always constant $a_0/2$ in a way that it converges to $s$. Maybe one can do someting with the distributive law but I am not sure if I am allowed to use it.
In particular if $L=sum (-1)^n frac1n+1$
then also
$L=frac12+frac12big(frac11cdot 2-frac12cdot 3+frac13cdot 4-frac14cdot 5+ …big)$
sequences-and-series convergence
sequences-and-series convergence
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
edited Mar 21 at 20:32
J. W. Tanner
4,4791320
4,4791320
asked Mar 21 at 20:02
New2MathNew2Math
16715
asked Mar 21 at 20:02
New2MathNew2Math
16715
asked Mar 21 at 20:02
New2MathNew2Math
16715
16715
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34
add a comment |
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Generalizing on my comment:
$$
beginalign
frac12a_0+frac12sum_k=0^n(-1)^k(a_k-a_k+1)
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=0^n(-1)^ka_k+1\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=1^n+1(-1)^k-1a_k\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k+frac12sum_k=1^n+1(-1)^ka_k\
&=sum_k=0^n(-1)^ka_k+frac12(-1)^n+1,a_n+1
endalign
$$
If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Generalizing on my comment:
$$
beginalign
frac12a_0+frac12sum_k=0^n(-1)^k(a_k-a_k+1)
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=0^n(-1)^ka_k+1\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=1^n+1(-1)^k-1a_k\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k+frac12sum_k=1^n+1(-1)^ka_k\
&=sum_k=0^n(-1)^ka_k+frac12(-1)^n+1,a_n+1
endalign
$$
If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
$endgroup$
add a comment |
$begingroup$
Generalizing on my comment:
$$
beginalign
frac12a_0+frac12sum_k=0^n(-1)^k(a_k-a_k+1)
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=0^n(-1)^ka_k+1\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=1^n+1(-1)^k-1a_k\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k+frac12sum_k=1^n+1(-1)^ka_k\
&=sum_k=0^n(-1)^ka_k+frac12(-1)^n+1,a_n+1
endalign
$$
If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
$endgroup$
add a comment |
$begingroup$
Generalizing on my comment:
$$
beginalign
frac12a_0+frac12sum_k=0^n(-1)^k(a_k-a_k+1)
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=0^n(-1)^ka_k+1\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=1^n+1(-1)^k-1a_k\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k+frac12sum_k=1^n+1(-1)^ka_k\
&=sum_k=0^n(-1)^ka_k+frac12(-1)^n+1,a_n+1
endalign
$$
If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
$endgroup$
Generalizing on my comment:
$$
beginalign
frac12a_0+frac12sum_k=0^n(-1)^k(a_k-a_k+1)
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=0^n(-1)^ka_k+1\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k-frac12sum_k=1^n+1(-1)^k-1a_k\
&=frac12a_0+frac12sum_k=0^n(-1)^ka_k+frac12sum_k=1^n+1(-1)^ka_k\
&=sum_k=0^n(-1)^ka_k+frac12(-1)^n+1,a_n+1
endalign
$$
If you are uncertain what you can use in an infinite sum, apply rules to a finite sum and then take the limit.
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
answered Mar 21 at 20:39
robjohn♦robjohn
270k27313641
270k27313641
add a comment |
add a comment |
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$begingroup$
Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
$endgroup$
– robjohn♦
Mar 21 at 20:09
$begingroup$
To see what is going on, take a look at the simpler example of taking the sum from $n=0$ to $n=3$: $frac12a_0+frac12(a_0-a_1)-frac12(a_1-a_2)+frac12(a_2-a_3)-frac12(a_3-a_4)=a_0-a_1+a_2-a_3+frac12a_4$
$endgroup$
– robjohn♦
Mar 21 at 20:25
$begingroup$
I believe that $a_n=10^-n$ shows that the proposition about the error of the accelerated series doesn't hold; that is, $|s-big(frac 1 2a_0+frac 1 2sum_n=0^infty(-1)^n(a_n-a_n+1)big)|leq a_k+1 a_k+2$
$endgroup$
– robjohn♦
Mar 21 at 21:34