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Let $\gamma=\lim_{n\to\infty} F(n)$ where $$F(n)=1+\frac{1}{2}+\frac{1}{3}+\cdots\frac{1}{n}-\ln(n)$$ (This is Euler's constant.) How can I calculate $\gamma$ with $10$ digits of precision using the Euler-Maclaurin Formula?

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The Euler-Maclaurin Formula says $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}\color{#C00}{-\frac1{132n^{10}}}+\dots $$ If we use $n=10$ and the expansion with and without the red term, we get that $$ 0.5772156649008\le\gamma\le0.5772156649016 $$ Thus, to $10$ places we get $$ \gamma=0.5772156649 $$

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  • $\begingroup$ Where does the $\gamma$ come from in $ \sum_{k=1}^n\frac1k=\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}\color{#C00}{-\frac1{132n^{10}}}+\dots $? $\endgroup$ – mathie12 Jun 17 '18 at 14:54
  • $\begingroup$ The definition of $$\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\log(n)\right)$$ $\endgroup$ – robjohn Jun 17 '18 at 15:18
  • $\begingroup$ I don't know if I understand, what I thought you did was: $ \sum_{k=1}^n\frac1k=\int_1^{n}\frac1xdx+\sum_{k=1}^p\frac{B_k}{k!}(f^{k-1}(n)-f^{k-1}(m))$ And doing this I get $ \sum_{k=1}^n\frac1k=\log(n)+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}+\frac{-1}{2}+\frac{1}{12}+\frac{1}{120}\dots$. What did I do wrong? $\endgroup$ – mathie12 Jun 17 '18 at 16:23
  • $\begingroup$ Essentially, the way I am using the Euler-Maclaurin Sum Formula is to estimate the tail of the sum. That is, using $m$ large and $n\to\infty$. If we try to use the formula for $m=1$ and large $n$, then the sum usually diverges, which makes it hard to use. $\endgroup$ – robjohn Jun 17 '18 at 19:32

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