Is the sequence of the even Bernoulli numbers $B_{2n}$ bounded? And if so is it posible to define a Dirichlet series $$f(s)= \sum_{n=1}^\infty \frac{a_n}{n^s}$$ with $a_n=B_{2n}$ that is convergent at $s=1$ or that one can assign a principal value in this point?
Bernoulli number sequence
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sequences-and-series
bernoulli-numbers
1 Answers
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$$B_{2n} \sim (-1)^{n-1} 4 \sqrt{\pi n} \left( \frac{n}{\pi e} \right)^{2n}$$ as can be seen from the formula for $\zeta (2n)$ and applying Sterling's formula, so the sequence of even Bernoulli numbers is unbounded.
I haven't seen a Dirichlet series for the Bernoulli numbers, but they do have an exponential generating function and a normal generating function if that helps. Another approach might be to take a look at Tao's blog post on smoothing sums to evaluate asymptotics of partial sums of divergent series. He uses it to find $-1/12$ as a constant in the expansion of $\sum_{n \leq N} n$ as well as to find values for the zeta function at other negative integers. That said, the Bernoulli numbers might grow too quickly to behave nicely (for instance, this technique fails for $\sum_{n \leq N} n!$).
EDIT: The digamma function has an asymptotic expansion of
$$\psi(x) = \ln x - \frac{1}{2x} + \sum_{n=1}^\infty \frac{B_{2n}}{2n x^{2n}}$$
that doesn't converge for any $x$, but is useful if you truncate the series to a finite number of terms. Plugging in $x=1$ to both sides yields
$$ \frac{1}{2} -\gamma = \sum_{n=1}^\infty \frac{B_{2n}}{2n}$$
though I'm not sure this makes any sense to consider.