A binomial double sumSumming a binomial seriesSum of squares of binomial coefficientsFinding the sum of a conditionally convergent double seriesDouble sum and zeta functionEvaluting sum $sum limits_n=0^inftyfracn^kn!$Does this summation (involving binomial) have a closed form? If so, what is it?Finding a closed form for $sum_k=1^infty(-1)^n-1fraczeta(2n+1)2^2n+1$Seeking general formula for Euler Sum $sumlimits _ n=1 ^ infty frac left(H_nright)^pn^q$ with $q$ even and $p$ oddHelp with a nasty sum of binomial coefficientsA Ramanujan sum
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A binomial double sum
Summing a binomial seriesSum of squares of binomial coefficientsFinding the sum of a conditionally convergent double seriesDouble sum and zeta functionEvaluting sum $sum limits_n=0^inftyfracn^kn!$Does this summation (involving binomial) have a closed form? If so, what is it?Finding a closed form for $sum_k=1^infty(-1)^n-1fraczeta(2n+1)2^2n+1$Seeking general formula for Euler Sum $sumlimits _ n=1 ^ infty frac left(H_nright)^pn^q$ with $q$ even and $p$ oddHelp with a nasty sum of binomial coefficientsA Ramanujan sum
$begingroup$
I was doing some numerical experiment and I found that
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n+1)=sum_k=2^inftyfrac1(k-1)k^2=2-zeta(2).$$
I wonder if the following variation
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)$$
has a closed form. Is it related to $zeta(2)$ or $zeta(3)$?
sequences-and-series
asked Mar 5 at 12:59
Frank SFrank S
637
$endgroup$
add a comment |
$begingroup$
I was doing some numerical experiment and I found that
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n+1)=sum_k=2^inftyfrac1(k-1)k^2=2-zeta(2).$$
I wonder if the following variation
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)$$
has a closed form. Is it related to $zeta(2)$ or $zeta(3)$?
sequences-and-series
asked Mar 5 at 12:59
Frank SFrank S
637
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
I was doing some numerical experiment and I found that
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n+1)=sum_k=2^inftyfrac1(k-1)k^2=2-zeta(2).$$
I wonder if the following variation
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)$$
has a closed form. Is it related to $zeta(2)$ or $zeta(3)$?
sequences-and-series
asked Mar 5 at 12:59
Frank SFrank S
637
$endgroup$
I was doing some numerical experiment and I found that
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n+1)=sum_k=2^inftyfrac1(k-1)k^2=2-zeta(2).$$
I wonder if the following variation
$$sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)$$
has a closed form. Is it related to $zeta(2)$ or $zeta(3)$?
sequences-and-series
sequences-and-series
asked Mar 5 at 12:59
Frank SFrank S
637
asked Mar 5 at 12:59
Frank SFrank S
637
edited Mar 5 at 13:35
Frank S
asked Mar 5 at 12:59
Frank SFrank S
637
asked Mar 5 at 12:59
Frank SFrank S
637
asked Mar 5 at 12:59
Frank SFrank S
637
637
$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday
add a comment |
$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday
$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday
$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have that the inner sum equals
$$
eqalign
& sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_0, le ,n 1 over binomn+kkleft( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + k right)^,underline ,k, left( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) cr
$$
Therefore
$$
eqalign
& S = sumlimits_2, le ,k 1 over kleft( k - 1 right)sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_2, le ,k k! over kleft( k - 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) = cr
& = sumlimits_0, le ,k left( k + 2 right)! over left( k + 2 right)left( k + 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k! over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k!,left( n + 1 right)^,overline ,k,
over left( n + 1 right)^,overline ,k + 2, left( n + 1 right)^,overline ,k + 1, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)sumlimits_0, le ,k
1^,overline ,k, ,left( n + 1 right)^,overline ,k, over left( n + 3 right)^,overline ,k, left( n + 2 right)^,overline ,k, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)
_3F_,2left( left. matrix 1,;1,n + 1 cr n + 3,n + 2 cr ; right right) cr
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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$begingroup$
We have that the inner sum equals
$$
eqalign
& sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_0, le ,n 1 over binomn+kkleft( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + k right)^,underline ,k, left( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) cr
$$
Therefore
$$
eqalign
& S = sumlimits_2, le ,k 1 over kleft( k - 1 right)sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_2, le ,k k! over kleft( k - 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) = cr
& = sumlimits_0, le ,k left( k + 2 right)! over left( k + 2 right)left( k + 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k! over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k!,left( n + 1 right)^,overline ,k,
over left( n + 1 right)^,overline ,k + 2, left( n + 1 right)^,overline ,k + 1, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)sumlimits_0, le ,k
1^,overline ,k, ,left( n + 1 right)^,overline ,k, over left( n + 3 right)^,overline ,k, left( n + 2 right)^,overline ,k, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)
_3F_,2left( left. matrix 1,;1,n + 1 cr n + 3,n + 2 cr ; right right) cr
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
add a comment |
$begingroup$
We have that the inner sum equals
$$
eqalign
& sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_0, le ,n 1 over binomn+kkleft( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + k right)^,underline ,k, left( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) cr
$$
Therefore
$$
eqalign
& S = sumlimits_2, le ,k 1 over kleft( k - 1 right)sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_2, le ,k k! over kleft( k - 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) = cr
& = sumlimits_0, le ,k left( k + 2 right)! over left( k + 2 right)left( k + 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k! over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k!,left( n + 1 right)^,overline ,k,
over left( n + 1 right)^,overline ,k + 2, left( n + 1 right)^,overline ,k + 1, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)sumlimits_0, le ,k
1^,overline ,k, ,left( n + 1 right)^,overline ,k, over left( n + 3 right)^,overline ,k, left( n + 2 right)^,overline ,k, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)
_3F_,2left( left. matrix 1,;1,n + 1 cr n + 3,n + 2 cr ; right right) cr
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
add a comment |
$begingroup$
We have that the inner sum equals
$$
eqalign
& sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_0, le ,n 1 over binomn+kkleft( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + k right)^,underline ,k, left( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) cr
$$
Therefore
$$
eqalign
& S = sumlimits_2, le ,k 1 over kleft( k - 1 right)sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_2, le ,k k! over kleft( k - 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) = cr
& = sumlimits_0, le ,k left( k + 2 right)! over left( k + 2 right)left( k + 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k! over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k!,left( n + 1 right)^,overline ,k,
over left( n + 1 right)^,overline ,k + 2, left( n + 1 right)^,overline ,k + 1, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)sumlimits_0, le ,k
1^,overline ,k, ,left( n + 1 right)^,overline ,k, over left( n + 3 right)^,overline ,k, left( n + 2 right)^,overline ,k, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)
_3F_,2left( left. matrix 1,;1,n + 1 cr n + 3,n + 2 cr ; right right) cr
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
$endgroup$
We have that the inner sum equals
$$
eqalign
& sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_0, le ,n 1 over binomn+kkleft( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + k right)^,underline ,k, left( n + k - 1 right) = cr
& = k!sumlimits_0, le ,n 1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) cr
$$
Therefore
$$
eqalign
& S = sumlimits_2, le ,k 1 over kleft( k - 1 right)sumlimits_k, le ,n 1 over binomnkleft( n - 1 right) = cr
& = sumlimits_2, le ,k k! over kleft( k - 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k, left( n + k - 1 right) = cr
& = sumlimits_0, le ,k left( k + 2 right)! over left( k + 2 right)left( k + 1 right)sumlimits_0, le ,n
1 over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k! over left( n + 1 right)^,overline ,k + 2, left( n + 1 + k right) = cr
& = sumlimits_0, le ,n sumlimits_0, le ,k k!,left( n + 1 right)^,overline ,k,
over left( n + 1 right)^,overline ,k + 2, left( n + 1 right)^,overline ,k + 1, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)sumlimits_0, le ,k
1^,overline ,k, ,left( n + 1 right)^,overline ,k, over left( n + 3 right)^,overline ,k, left( n + 2 right)^,overline ,k, = cr
& = sumlimits_0, le ,n 1 over left( n + 1 right)^,2 left( n + 2 right)
_3F_,2left( left. matrix 1,;1,n + 1 cr n + 3,n + 2 cr ; right right) cr
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
edited yesterday
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
answered Mar 5 at 16:06
G CabG Cab
19.9k31340
19.9k31340
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
add a comment |
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
Are you saying that the given series is divergent? Note that $sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk(n-1)leq sum_k=2^inftyfrac1k(k-1)sum_n=k^inftyfrac1binomnk=sum_k=2^inftyfrac1(k-1)^2=zeta(2)$.
$endgroup$
– Robert Z
Mar 5 at 16:22
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
$begingroup$
@RobertZ: ehps, probably I made some error: will verify, thanks for signalling
$endgroup$
– G Cab
Mar 5 at 17:16
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– clathratus
yesterday