Partial summation formula for two indexes?
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We know that partial summation states that given two sequences $(alpha_n), (beta_n)$, we can write
$$sum_{n=1}^palpha_nbeta_n=beta_nA_p+sum_{n=1}^{p-1}(beta_n-beta_{n+1})A_n$$
where $A_p=sum_{n=1}^{p}alpha_n$.
Is there a version of this formula for sequences $(alpha_n^m)$ with two indexes?
sequences-and-series analysis summation-by-parts
asked Nov 16 at 20:00
Mark_Hoffman
663515
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up vote
0
down vote
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We know that partial summation states that given two sequences $(alpha_n), (beta_n)$, we can write
$$sum_{n=1}^palpha_nbeta_n=beta_nA_p+sum_{n=1}^{p-1}(beta_n-beta_{n+1})A_n$$
where $A_p=sum_{n=1}^{p}alpha_n$.
Is there a version of this formula for sequences $(alpha_n^m)$ with two indexes?
sequences-and-series analysis summation-by-parts
asked Nov 16 at 20:00
Mark_Hoffman
663515
1
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We know that partial summation states that given two sequences $(alpha_n), (beta_n)$, we can write
$$sum_{n=1}^palpha_nbeta_n=beta_nA_p+sum_{n=1}^{p-1}(beta_n-beta_{n+1})A_n$$
where $A_p=sum_{n=1}^{p}alpha_n$.
Is there a version of this formula for sequences $(alpha_n^m)$ with two indexes?
sequences-and-series analysis summation-by-parts
asked Nov 16 at 20:00
Mark_Hoffman
663515
We know that partial summation states that given two sequences $(alpha_n), (beta_n)$, we can write
$$sum_{n=1}^palpha_nbeta_n=beta_nA_p+sum_{n=1}^{p-1}(beta_n-beta_{n+1})A_n$$
where $A_p=sum_{n=1}^{p}alpha_n$.
Is there a version of this formula for sequences $(alpha_n^m)$ with two indexes?
sequences-and-series analysis summation-by-parts
sequences-and-series analysis summation-by-parts
asked Nov 16 at 20:00
Mark_Hoffman
663515
asked Nov 16 at 20:00
Mark_Hoffman
663515
edited Nov 16 at 21:37
asked Nov 16 at 20:00
Mark_Hoffman
663515
asked Nov 16 at 20:00
Mark_Hoffman
663515
asked Nov 16 at 20:00
Mark_Hoffman
663515
663515
1
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38
add a comment |
1
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38
1
1
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38
add a comment |
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1
Do you sum over $m$ as well?
– Andrei
Nov 16 at 20:05
Yes, I'm thinking about $sum_{n,m}alpha_n^mbeta_n^m$
– Mark_Hoffman
Nov 16 at 21:38