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On of methods for value of an infinite sum like $\zeta (3)$ is using following formula which by this trick we can write an infinite sum to a faster sum which by summing first terms of it we can find decimal digits .

$<math>\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)</math>$$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ and by taking $f(n)=\frac{1}{n^3}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

But the second method which is more welcomed for number theorist which working on $\zeta (3)$ in recent decade is generationg function method. In fact Bernoulli numbers which is very important in generating function method is less functional for $\zeta (3)$ . So by this reason Kaneko defined a new generating function which was more applicable for finding $\zeta (3)$ up to now.

$<math>{Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}</math>$${Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}$

where Li is the polylogarithm. The $<math>B_{n}^{(1)}</math> $$B_{n}^{(1)} $are the usual Bernoulli numbers.

Kaneko also gave following combinatorial formula:

$<math>B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},</math>$$B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},$

You can follow following paper. So you can write Euler-Maclaurin formula with respect to $B_n^{(k)}$ and get more desired results for decimal digits of $\zeta (3)$.

On of methods for value of an infinite sum like $\zeta (3)$ is using following formula which by this trick we can write an infinite sum to a faster sum which by summing first terms of it we can find decimal digits .

$<math>\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)</math>$ and by taking $f(n)=\frac{1}{n^3}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

But the second method which is more welcomed for number theorist which working on $\zeta (3)$ in recent decade is generationg function method. In fact Bernoulli numbers which is very important in generating function method is less functional for $\zeta (3)$ . So by this reason Kaneko defined a new generating function which was more applicable for finding $\zeta (3)$ up to now.

$<math>{Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}</math>$

where Li is the polylogarithm. The $<math>B_{n}^{(1)}</math> $are the usual Bernoulli numbers.

Kaneko also gave following combinatorial formula:

$<math>B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},</math>$

You can follow following paper. So you can write Euler-Maclaurin formula with respect to $B_n^{(k)}$ and get more desired results for decimal digits of $\zeta (3)$.

On of methods for value of an infinite sum like $\zeta (3)$ is using following formula which by this trick we can write an infinite sum to a faster sum which by summing first terms of it we can find decimal digits .

$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ and by taking $f(n)=\frac{1}{n^3}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

But the second method which is more welcomed for number theorist which working on $\zeta (3)$ in recent decade is generationg function method. In fact Bernoulli numbers which is very important in generating function method is less functional for $\zeta (3)$ . So by this reason Kaneko defined a new generating function which was more applicable for finding $\zeta (3)$ up to now.

${Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}$

where Li is the polylogarithm. The $B_{n}^{(1)} $are the usual Bernoulli numbers.

Kaneko also gave following combinatorial formula:

$B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},$

You can follow following paper. So you can write Euler-Maclaurin formula with respect to $B_n^{(k)}$ and get more desired results for decimal digits of $\zeta (3)$.

Source Link
user21574
user21574

On of methods for value of an infinite sum like $\zeta (3)$ is using following formula which by this trick we can write an infinite sum to a faster sum which by summing first terms of it we can find decimal digits .

$<math>\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)</math>$ and by taking $f(n)=\frac{1}{n^3}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

But the second method which is more welcomed for number theorist which working on $\zeta (3)$ in recent decade is generationg function method. In fact Bernoulli numbers which is very important in generating function method is less functional for $\zeta (3)$ . So by this reason Kaneko defined a new generating function which was more applicable for finding $\zeta (3)$ up to now.

$<math>{Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}</math>$

where Li is the polylogarithm. The $<math>B_{n}^{(1)}</math> $are the usual Bernoulli numbers.

Kaneko also gave following combinatorial formula:

$<math>B_{n}^{(-k)}=\sum_{m=0}^{n}(-1)^{m+n}m!S(n,m)(m+1)^{k},</math>$

You can follow following paper. So you can write Euler-Maclaurin formula with respect to $B_n^{(k)}$ and get more desired results for decimal digits of $\zeta (3)$.