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.
Best 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, $$ \left| {R_N (n)} \right| \le \frac{{\left| {B_{2N} } \right|}}{{2N}}\frac{1}{{n^{2N} }} = \frac{{(2N)!}}{{(2\pi n)^{2N} N}}\zeta (2N) = \mathcal{O}\!\left( {\frac{{\mathrm{e}^{ - 2\pi n} }}{{\sqrt n }}} \right). $$ 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 $\log n!$.