Mixing
Iterated Sums of Differences
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I recently came across the sum $$sum_n=0^infty (e- S_n) = e,$$ where $S_n = 1+1+1/2+1/6+...+1/n!$ is the classic $n$th partial sum of $e.$ This gave me an idea. Let $a_0,0, a_0,1, a_0,2,dots$ be a convergent sequence, let $$S_m,n=sum_k=n^infty a_m,k$$
$$a_i+1,j = S_i,j+1$$
Now let $b_n=S_n,0.$ For the sequence $a_0,n=1/n!,$ we have $b_0=e,$ and as I mentioned earlier, $b_1=e.$ I think that $1<b_2<2,$ as the sum decays quickly after the initial term. Also interesting is the fact that if you take $a_0,n=0.5^n,$ then $b_n=2 , forall n,$ as $a_i,j=a_0,j=2^-j , forall i,j.$ Perhaps this is the only such sequence with the property that $b_n$ is constant.
Is there any merit to such sequences? I would like to know if it is possible to determine the long term behavior of $b_n$ for various sequences $a_0,n.$ I believe that someone has already thought of this before, and done extensive research on it, as such an idea seems too simple to have been overlooked.
sequences-and-series recurrence-relations recursion
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I recently came across the sum $$sum_n=0^infty (e- S_n) = e,$$ where $S_n = 1+1+1/2+1/6+...+1/n!$ is the classic $n$th partial sum of $e.$ This gave me an idea. Let $a_0,0, a_0,1, a_0,2,dots$ be a convergent sequence, let $$S_m,n=sum_k=n^infty a_m,k$$
$$a_i+1,j = S_i,j+1$$
Now let $b_n=S_n,0.$ For the sequence $a_0,n=1/n!,$ we have $b_0=e,$ and as I mentioned earlier, $b_1=e.$ I think that $1<b_2<2,$ as the sum decays quickly after the initial term. Also interesting is the fact that if you take $a_0,n=0.5^n,$ then $b_n=2 , forall n,$ as $a_i,j=a_0,j=2^-j , forall i,j.$ Perhaps this is the only such sequence with the property that $b_n$ is constant.
Is there any merit to such sequences? I would like to know if it is possible to determine the long term behavior of $b_n$ for various sequences $a_0,n.$ I believe that someone has already thought of this before, and done extensive research on it, as such an idea seems too simple to have been overlooked.
sequences-and-series recurrence-relations recursion
asked Jul 15 at 0:54
Display name
623211
Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12
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I recently came across the sum $$sum_n=0^infty (e- S_n) = e,$$ where $S_n = 1+1+1/2+1/6+...+1/n!$ is the classic $n$th partial sum of $e.$ This gave me an idea. Let $a_0,0, a_0,1, a_0,2,dots$ be a convergent sequence, let $$S_m,n=sum_k=n^infty a_m,k$$
$$a_i+1,j = S_i,j+1$$
Now let $b_n=S_n,0.$ For the sequence $a_0,n=1/n!,$ we have $b_0=e,$ and as I mentioned earlier, $b_1=e.$ I think that $1<b_2<2,$ as the sum decays quickly after the initial term. Also interesting is the fact that if you take $a_0,n=0.5^n,$ then $b_n=2 , forall n,$ as $a_i,j=a_0,j=2^-j , forall i,j.$ Perhaps this is the only such sequence with the property that $b_n$ is constant.
Is there any merit to such sequences? I would like to know if it is possible to determine the long term behavior of $b_n$ for various sequences $a_0,n.$ I believe that someone has already thought of this before, and done extensive research on it, as such an idea seems too simple to have been overlooked.
sequences-and-series recurrence-relations recursion
asked Jul 15 at 0:54
Display name
623211
I recently came across the sum $$sum_n=0^infty (e- S_n) = e,$$ where $S_n = 1+1+1/2+1/6+...+1/n!$ is the classic $n$th partial sum of $e.$ This gave me an idea. Let $a_0,0, a_0,1, a_0,2,dots$ be a convergent sequence, let $$S_m,n=sum_k=n^infty a_m,k$$
$$a_i+1,j = S_i,j+1$$
Now let $b_n=S_n,0.$ For the sequence $a_0,n=1/n!,$ we have $b_0=e,$ and as I mentioned earlier, $b_1=e.$ I think that $1<b_2<2,$ as the sum decays quickly after the initial term. Also interesting is the fact that if you take $a_0,n=0.5^n,$ then $b_n=2 , forall n,$ as $a_i,j=a_0,j=2^-j , forall i,j.$ Perhaps this is the only such sequence with the property that $b_n$ is constant.
Is there any merit to such sequences? I would like to know if it is possible to determine the long term behavior of $b_n$ for various sequences $a_0,n.$ I believe that someone has already thought of this before, and done extensive research on it, as such an idea seems too simple to have been overlooked.
sequences-and-series recurrence-relations recursion
asked Jul 15 at 0:54
Display name
623211
edited Jul 15 at 1:01
asked Jul 15 at 0:54
Display name
623211
asked Jul 15 at 0:54
Display name
623211
asked Jul 15 at 0:54
Display name
623211
623211
Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12
add a comment |Â
Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12
Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12
add a comment |Â
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Looks wrong. Should be infinity.
– Moti
Jul 15 at 6:03
Perhaps an example is better. Consider the sequence 1,1/2,1/4,1/8...; it has sum 2, so form the next series 2-1, 2-(1+1/2), 2-(1+1/2+1/4),... which is identical to the original series and thus also has sum 2. This repeats ad infinitum, explaining why b_n=2 for all n in this case. Now for my other example, consider the sequence 1,1,1/2,1/6,1/24...; it has sum $e,$ so we form the series $e-1, e-(1+1), e-(1+1+1/2),...$ which remarkably also has sum $e.$ Now the next series is $e-(e-1)=1, e-(e-1+e-2)=3-e, e-(e-1+e-2+e-2.5)=5.5-2e,...$ which seems to have $1<S<2;$ I haven't found an exact value.
– Display name
Jul 16 at 18:12