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?


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 $$ \gamma=0.5772156649 $$

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.