Evaluating $frac1{3^k}+frac1{6^k}+frac1{10^k}+cdots$
$begingroup$
The series $frac1{3^k}+frac1{6^k}+frac1{10^k}+cdots$ for integers $k>1$ at the triangular numbers can be written as $$sum_{i=2}^inftyfrac1{left(frac{i(i+1)}2right)^k}=2^ksum_{i=2}^inftyfrac1{(i(i+1))^k}$$ At first glance, the zeta function may be of use. Partial fractions can be used to evaluate the sum, but only for small values of $k$ since the general formula is $$frac1{i^k(i+1)^k}=frac{a_1}i+frac{a_2}{i^2}+cdots+frac{a_k}{i^k}+frac{b_1}{i+1}+frac{b_2}{(i+1)^2}+cdots+frac{b_k}{(i+1)^k}.$$ Of course, once the expression in the sum is presented as such, $zeta(cdot)$ can replace each individual term. However, apart from this tedious method, I do not see a way to get rid of the summation.
Is there another way?
sequences-and-series riemann-zeta
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
add a comment |
$begingroup$
The series $frac1{3^k}+frac1{6^k}+frac1{10^k}+cdots$ for integers $k>1$ at the triangular numbers can be written as $$sum_{i=2}^inftyfrac1{left(frac{i(i+1)}2right)^k}=2^ksum_{i=2}^inftyfrac1{(i(i+1))^k}$$ At first glance, the zeta function may be of use. Partial fractions can be used to evaluate the sum, but only for small values of $k$ since the general formula is $$frac1{i^k(i+1)^k}=frac{a_1}i+frac{a_2}{i^2}+cdots+frac{a_k}{i^k}+frac{b_1}{i+1}+frac{b_2}{(i+1)^2}+cdots+frac{b_k}{(i+1)^k}.$$ Of course, once the expression in the sum is presented as such, $zeta(cdot)$ can replace each individual term. However, apart from this tedious method, I do not see a way to get rid of the summation.
Is there another way?
sequences-and-series riemann-zeta
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
1
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27
add a comment |
$begingroup$
The series $frac1{3^k}+frac1{6^k}+frac1{10^k}+cdots$ for integers $k>1$ at the triangular numbers can be written as $$sum_{i=2}^inftyfrac1{left(frac{i(i+1)}2right)^k}=2^ksum_{i=2}^inftyfrac1{(i(i+1))^k}$$ At first glance, the zeta function may be of use. Partial fractions can be used to evaluate the sum, but only for small values of $k$ since the general formula is $$frac1{i^k(i+1)^k}=frac{a_1}i+frac{a_2}{i^2}+cdots+frac{a_k}{i^k}+frac{b_1}{i+1}+frac{b_2}{(i+1)^2}+cdots+frac{b_k}{(i+1)^k}.$$ Of course, once the expression in the sum is presented as such, $zeta(cdot)$ can replace each individual term. However, apart from this tedious method, I do not see a way to get rid of the summation.
Is there another way?
sequences-and-series riemann-zeta
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
$endgroup$
The series $frac1{3^k}+frac1{6^k}+frac1{10^k}+cdots$ for integers $k>1$ at the triangular numbers can be written as $$sum_{i=2}^inftyfrac1{left(frac{i(i+1)}2right)^k}=2^ksum_{i=2}^inftyfrac1{(i(i+1))^k}$$ At first glance, the zeta function may be of use. Partial fractions can be used to evaluate the sum, but only for small values of $k$ since the general formula is $$frac1{i^k(i+1)^k}=frac{a_1}i+frac{a_2}{i^2}+cdots+frac{a_k}{i^k}+frac{b_1}{i+1}+frac{b_2}{(i+1)^2}+cdots+frac{b_k}{(i+1)^k}.$$ Of course, once the expression in the sum is presented as such, $zeta(cdot)$ can replace each individual term. However, apart from this tedious method, I do not see a way to get rid of the summation.
Is there another way?
sequences-and-series riemann-zeta
sequences-and-series riemann-zeta
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
asked Dec 22 '18 at 15:52
TheSimpliFireTheSimpliFire
13.2k62464
13.2k62464
$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
1
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27
add a comment |
$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
1
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27
$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
1
1
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just a few values.
$$left(
begin{array}{cc}
1 & 1 \
2 & -13+frac{4 pi ^2}{3} \
3 & 79-8 pi ^2 \
4 & -561+frac{160 pi ^2}{3}+frac{16 pi ^4}{45} \
5 & 4031-frac{1120 pi ^2}{3}-frac{32 pi ^4}{9} \
6 & -29569+2688 pi ^2+frac{448 pi ^4}{15}+frac{128 pi ^6}{945} \
7 & 219647-19712 pi ^2-frac{3584 pi ^4}{15}-frac{256 pi ^6}{135} \
8 & -1647361+146432 pi ^2+frac{5632 pi ^4}{3}+frac{2048 pi ^6}{105}+frac{256
pi ^8}{4725} \
9 & 12446719-1098240 pi ^2-frac{73216 pi ^4}{5}-frac{11264 pi
^6}{63}-frac{512 pi ^8}{525} \
10 & -94595073+frac{24893440 pi ^2}{3}+frac{1025024 pi ^4}{9}+frac{292864 pi
^6}{189}+frac{11264 pi ^8}{945}+frac{2048 pi ^{10}}{93555}
end{array}
right)$$
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
Just a few values.
$$left(
begin{array}{cc}
1 & 1 \
2 & -13+frac{4 pi ^2}{3} \
3 & 79-8 pi ^2 \
4 & -561+frac{160 pi ^2}{3}+frac{16 pi ^4}{45} \
5 & 4031-frac{1120 pi ^2}{3}-frac{32 pi ^4}{9} \
6 & -29569+2688 pi ^2+frac{448 pi ^4}{15}+frac{128 pi ^6}{945} \
7 & 219647-19712 pi ^2-frac{3584 pi ^4}{15}-frac{256 pi ^6}{135} \
8 & -1647361+146432 pi ^2+frac{5632 pi ^4}{3}+frac{2048 pi ^6}{105}+frac{256
pi ^8}{4725} \
9 & 12446719-1098240 pi ^2-frac{73216 pi ^4}{5}-frac{11264 pi
^6}{63}-frac{512 pi ^8}{525} \
10 & -94595073+frac{24893440 pi ^2}{3}+frac{1025024 pi ^4}{9}+frac{292864 pi
^6}{189}+frac{11264 pi ^8}{945}+frac{2048 pi ^{10}}{93555}
end{array}
right)$$
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Just a few values.
$$left(
begin{array}{cc}
1 & 1 \
2 & -13+frac{4 pi ^2}{3} \
3 & 79-8 pi ^2 \
4 & -561+frac{160 pi ^2}{3}+frac{16 pi ^4}{45} \
5 & 4031-frac{1120 pi ^2}{3}-frac{32 pi ^4}{9} \
6 & -29569+2688 pi ^2+frac{448 pi ^4}{15}+frac{128 pi ^6}{945} \
7 & 219647-19712 pi ^2-frac{3584 pi ^4}{15}-frac{256 pi ^6}{135} \
8 & -1647361+146432 pi ^2+frac{5632 pi ^4}{3}+frac{2048 pi ^6}{105}+frac{256
pi ^8}{4725} \
9 & 12446719-1098240 pi ^2-frac{73216 pi ^4}{5}-frac{11264 pi
^6}{63}-frac{512 pi ^8}{525} \
10 & -94595073+frac{24893440 pi ^2}{3}+frac{1025024 pi ^4}{9}+frac{292864 pi
^6}{189}+frac{11264 pi ^8}{945}+frac{2048 pi ^{10}}{93555}
end{array}
right)$$
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Just a few values.
$$left(
begin{array}{cc}
1 & 1 \
2 & -13+frac{4 pi ^2}{3} \
3 & 79-8 pi ^2 \
4 & -561+frac{160 pi ^2}{3}+frac{16 pi ^4}{45} \
5 & 4031-frac{1120 pi ^2}{3}-frac{32 pi ^4}{9} \
6 & -29569+2688 pi ^2+frac{448 pi ^4}{15}+frac{128 pi ^6}{945} \
7 & 219647-19712 pi ^2-frac{3584 pi ^4}{15}-frac{256 pi ^6}{135} \
8 & -1647361+146432 pi ^2+frac{5632 pi ^4}{3}+frac{2048 pi ^6}{105}+frac{256
pi ^8}{4725} \
9 & 12446719-1098240 pi ^2-frac{73216 pi ^4}{5}-frac{11264 pi
^6}{63}-frac{512 pi ^8}{525} \
10 & -94595073+frac{24893440 pi ^2}{3}+frac{1025024 pi ^4}{9}+frac{292864 pi
^6}{189}+frac{11264 pi ^8}{945}+frac{2048 pi ^{10}}{93555}
end{array}
right)$$
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
Just a few values.
$$left(
begin{array}{cc}
1 & 1 \
2 & -13+frac{4 pi ^2}{3} \
3 & 79-8 pi ^2 \
4 & -561+frac{160 pi ^2}{3}+frac{16 pi ^4}{45} \
5 & 4031-frac{1120 pi ^2}{3}-frac{32 pi ^4}{9} \
6 & -29569+2688 pi ^2+frac{448 pi ^4}{15}+frac{128 pi ^6}{945} \
7 & 219647-19712 pi ^2-frac{3584 pi ^4}{15}-frac{256 pi ^6}{135} \
8 & -1647361+146432 pi ^2+frac{5632 pi ^4}{3}+frac{2048 pi ^6}{105}+frac{256
pi ^8}{4725} \
9 & 12446719-1098240 pi ^2-frac{73216 pi ^4}{5}-frac{11264 pi
^6}{63}-frac{512 pi ^8}{525} \
10 & -94595073+frac{24893440 pi ^2}{3}+frac{1025024 pi ^4}{9}+frac{292864 pi
^6}{189}+frac{11264 pi ^8}{945}+frac{2048 pi ^{10}}{93555}
end{array}
right)$$
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
answered Dec 22 '18 at 16:31
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
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$begingroup$
Why start at i=2 and not at 1?
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 15:58
$begingroup$
It might help if you solve first for specific k, like k=1 and k=2.
$endgroup$
– ypercubeᵀᴹ
Dec 22 '18 at 16:01
1
$begingroup$
@yper-crazyhat-cubeᵀᴹ $k=1$ is trivial. For $k=2,3,4$, the sums are $-frac{13}4+2zeta(2)$, $-frac{79}8-6zeta(2)$ and $-frac{561}{16}+20zeta(2)+2zeta(4)$ respectively. Looks like a negative coefficient, then terms of $zeta(2s)$ for $sge 1$.
$endgroup$
– TheSimpliFire
Dec 22 '18 at 16:02
$begingroup$
From the first values for $k$ which you gave, it appears that the result will be a weighted sum of $zeta(2 n) $ with $n = 0,1,cdots,lfloor k/2 rfloor $, where the weights are integer fractions. Further, it is known that $zeta(2 n) = frac{p}{q} pi^{2n}$ where $p,q $ are integers. [continued below]
$endgroup$
– Andreas
Dec 22 '18 at 16:26
$begingroup$
[continued from above] So, if the trend for the result holds, the general result will be of the form $sum_{i=0}^{lfloor k/2 rfloor} frac{p_i}{q_i} pi^{2i} $ where $p_i,q_i $ are integers. Since $pi$ is irrational, this could only be simplified if it happens to be a perfect power in $pi^2$, i.e. $a (pi^2 + b)^{lfloor k/2 rfloor}$, or at least partially so, with some lower power. It seems that this is not the case, which would result in the fact that indeed you need some fractions expansion with at least $lfloor k/2 rfloor$ many terms.
$endgroup$
– Andreas
Dec 22 '18 at 16:27