Prove that $sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that $sum_kge 1 k/(k+1)! = 1$how to prove that $sum_k=1^mk!k=(m+1)!-1$ without induction?How to prove that $sum_k=0^n binom2n+12k+1 = 4^n$Prove that $1.49<sum_k=1^99frac1k^2<1.99$Looking for a clever simplification of $sum_j=2^T_L(j-1-O_L)(2j-n-1)+sum_i=T^U^n-1(O^U-n+i)(2i-n-1)$What is the asymptotic behaviour of $sum_p_kleq xkp_k$, where $p_k$ is the kth prime number?How to prove that $sum_i=j^nn-i = sum_i=1^n-ji$?Prove that $ sum_x=1^ infty frac1x! = e -1 $Convergence of $sum_k=1^infty p_k^-z$, $p$ is prime.Tools for finding bounds on power series
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Prove that $sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that $sum_kge 1 k/(k+1)! = 1$how to prove that $sum_k=1^mk!k=(m+1)!-1$ without induction?How to prove that $sum_k=0^n binom2n+12k+1 = 4^n$Prove that $1.49<sum_k=1^99frac1k^2<1.99$Looking for a clever simplification of $sum_j=2^T_L(j-1-O_L)(2j-n-1)+sum_i=T^U^n-1(O^U-n+i)(2i-n-1)$What is the asymptotic behaviour of $sum_p_kleq xkp_k$, where $p_k$ is the kth prime number?How to prove that $sum_i=j^nn-i = sum_i=1^n-ji$?Prove that $ sum_x=1^ infty frac1x! = e -1 $Convergence of $sum_k=1^infty p_k^-z$, $p$ is prime.Tools for finding bounds on power series
$begingroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
$endgroup$
add a comment |
$begingroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
$endgroup$
add a comment |
$begingroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
$endgroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_j=0^n sum_k=0^j p_k q_j-k r_n-j = sum_k=0^n sum_j=k^n p_k q_j-k r_n-j$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2201621
1,2201621
add a comment |
add a comment |
2 Answers
2
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$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S 0leq j,lleq nland kleq j$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.7k394162
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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active
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$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
add a comment |
$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
add a comment |
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S 0leq j,lleq nland kleq j$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.7k394162
$endgroup$
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S 0leq j,lleq nland kleq j$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.7k394162
$endgroup$
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S 0leq j,lleq nland kleq j$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.7k394162
$endgroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S 0leq j,lleq nland kleq j$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.7k394162
answered Mar 26 at 10:22
5xum5xum
92.7k394162
answered Mar 26 at 10:22
5xum5xum
92.7k394162
answered Mar 26 at 10:22
5xum5xum
92.7k394162
92.7k394162
add a comment |
add a comment |
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