I'd like to find an asymptotic development for $\sum_{k=2}^{n} \ln(k)$
I know this sum is equivalent to $n\ln(n)-n$ but I cannot find more terms.
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1Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 03 '22 at 10:13
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Look into the Euler-Maclaurin Sum Formula, but do look at the links mentioned in José's comment. – robjohn Sep 03 '22 at 11:07
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Using Mathematica: $\frac{1}{1188 n^9}-\frac{1}{1680 n^7}+\frac{1}{1260 n^5}-\frac{1}{360 n^3}-n+\frac{1}{12 n}+\left(n+\frac{1}{2}\right) \ln (n)+\frac{1}{2} \ln (2 \pi )$ – Mariusz Iwaniuk Sep 03 '22 at 11:59
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1Probably, the shortcut is to use the Stirling's approximation $\displaystyle n!=\sqrt{2\pi n}\Big(\frac{n}{e}\Big)^n\Big(1+\frac{1}{12n}+...\Big)$ and notice that $\displaystyle \sum_{k=2}^n\ln k=\sum_{k=1}^n\ln k=\ln (n!)$ – Svyatoslav Sep 03 '22 at 12:04
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2This might be a duplicate of Summation of logs – robjohn Sep 03 '22 at 12:18
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Is it possible to do so by using the following inequality: ln (k)<= $ \int_{k}^(k+1) (ln(x)dx) $ <= ln(k+1) – alexandre rogojan Sep 03 '22 at 13:47