# Convergence of Bernoulli numbers infinite sum

In https://en.wikipedia.org/wiki/Harmonic_number I found that harmonic numbers admit asymptotic expansion as:

$$H_n \approx \ln n + \gamma_0 + \frac{1}{2n} - \sum_{k=1}^{\infty}{\frac{B_{2k}}{2kn^{2k}}}$$

but for a certain n, when trying to get some values on infinite sum on RHS it appears to diverge always, even if I try Cesaro sums pairing consecutive terms.

What is the real meaning of that infinite sum and how can I evaluate it?

Additionaly I am interested in some other infinite sums (in same depicted expression) that could formally converge without any exotic sumation argued.

• At a glance, it looks like the Euler-Maclaurin formula Dec 27 '20 at 17:18
• an asymptotic expansion doesn't need to converge (indeed generally it doesn't), see here Dec 27 '20 at 17:52

## 1 Answer

This is an example of an asymptotic power series. Its meaning is that for each $$N\geq 1$$, there exist $$A_N>0$$ and $$C_N>0$$ such that $$H_n = \log n + \gamma + \frac{1}{{2n}} - \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{2k}}\frac{1}{{n^{2k} }}} + R_N (n),$$ where $$\left| {R_N (n)} \right| \le \frac{{C_N }}{{n^{2N} }},$$ whenever $$n\geq A_N$$. In particular, $$R_N (n)=\mathcal{O}\!\left( {\frac{1}{{n^{2N} }}} \right)$$ as $$n\to +\infty$$. In this specific example, more can be said: for any $$n\geq 1$$ and $$N\geq 1$$, $$\left| {R_N (n)} \right| \le \frac{{|B_{2N}| }}{{2N}}\frac{1}{{n^{2N} }}.$$ Note that $$\left| {B_{2k} } \right| = \frac{{2(2k)!}}{{(2\pi )^{2k} }}\zeta (2k) > \frac{{2(2k)!}}{{(2\pi )^{2k} }},$$ so the terms of the series do not tend to $$0$$ for any $$n\geq 1$$ as $$k\to +\infty$$, i.e., the series cannot converge. Note however, that for each fixed $$n$$, the terms of the series initially decrease in absolute value and reach a minimum around $$k \approx \pi n$$. If the series is truncated at this optimal point, i.e., $$N = \left\lfloor {\pi n} \right\rfloor$$, then the error term satisfies, using Stirling's approximation for the factorial, $$\left| {R_N (n)} \right| \le \frac{{\left| {B_{2N} } \right|}}{{2N}}\frac{1}{{n^{2N} }} = \frac{{(2N)!}}{{(2\pi n)^{2N} N}}\zeta (2N) = \mathcal{O}\!\left( {\frac{{e^{ - 2\pi n} }}{{\sqrt n }}} \right).$$ This is called superasymptotics: the error is exponentially small. It is possible to re-expand the remainder into a new series which will improve the accuracy beyond superasymptotics. These are called hyperasymptotic expansions. In this particular example, the re-expansion will be a convergent series in terms of incomplete gamma functions which "terminates" the divergent asymptotic series. For a similar example of an asymptotic series like this, consider the Stirling series for $$\log n!$$.