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Bernoulli Numbers
- Thread starter PaulDirac
- Start date
As the primes. the Bernoulli numbers are infinite. How can we find the sum of all of them?.
That is, the denominators are smaller than the numerators and they alternate in sign.
See here: http://www.bernoulli.org/
i.e.\(\displaystyle B_{22}=\frac{854513}{138}\approx 6192\)
Though, as you probably know, the odd Bernoulli numbers are 0.
Here is a fun identity.
\(\displaystyle B_{2n}=(-1)^{n+1}\frac{2(2n)!}{(2{\pi})^{2n}}\left(1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+...\right)\)
That is, the denominators are smaller than the numerators and they alternate in sign.
See here: http://www.bernoulli.org/
i.e.\(\displaystyle B_{22}=\frac{854513}{138}\approx 6192\)
Though, as you probably know, the odd Bernoulli numbers are 0.
Here is a fun identity.
\(\displaystyle B_{2n}=(-1)^{n+1}\frac{2(2n)!}{(2{\pi})^{2n}}\left(1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+...\right)\)
The equation x/(e^(x)-1)=sum(n=0 to infty, (B(n)*(x)^n)/n!), where B(n) are the Bernoulli numbers, can be simplified by evaluating sum(n=0 to infty, 1/n!), which equals e. Then, by choosing a suitable value of x you can solve for the sum of all Bernoulli numbers. Letting x=1/2 yields .5/(e^(.5)-1)=e*sum(n=0 to infty, B(n)*(.5)^n)). sum(n=0 to infty, (.5)^n) can be evaluated using geometric sum rules to give 2. Therefore, .5/(e^(.5)-1)=2*e*sum(n=0 to infty, B(n)). Dividing both sides by 2*e and simplifying the left side gives .1417709954=sum(n=0 to infty, B(n)).