Consider the series $$ \sum_{k=2}^\infty \frac{\ln(k)}{k^p}, $$ which is easily seen to converge if $p>1$. Numerical computations seems to reveal that, if $n\in\mathbb N$ $$ \left\lceil\sum_{k=2}^\infty \frac{\ln(k)}{k^{1+10^{-n}}}\right\rceil=10^{2n}. $$ Is there an easy way to prove this?
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By definition of $\zeta$ function, you have$$S_n = \sum_{k=2}^\infty \frac{\ln(k)}{k^{1+10^{-n}}}=- \zeta'\left( 1+\frac{1}{10^n}\right)$$
Moreover, the series of $\zeta'$ near $1$ is given by $$\zeta'(s)=-\frac{1}{(s-1)^2}-\sum_{k=0}^{+\infty} \frac{(-1)^k}{k!} \gamma_{k+1} (s-1)^{k} $$
where the $\gamma_k$ are the Stieltjes constants
So $$S_n = 10^{2n} + \sum_{k=0}^{+\infty} \frac{(-1)^k}{k!} \gamma_{k+1} 10^{-nk} $$
and using the majoration $|\gamma_p| \leq \left( \frac{p}{e}\right)^p$, you can easily see that $$0 < \sum_{k=0}^{+\infty} \frac{(-1)^k}{k!} \gamma_{k+1} 10^{-nk}< 1$$
which shows that $$\left\lceil S_n \right\rceil = 10^{2n}$$
TheSilverDoe
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You mean the series without the ceiling, right? Do you have a reference for this identity? – frog May 04 '21 at 07:50
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@frog I edited ! – TheSilverDoe May 04 '21 at 08:15
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This is a very nice answer, thank you very much! – frog May 04 '21 at 11:06