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.
 A: 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 $\log n!$.
