Constante of Euler-Mascheroni

Question

I know that $$\gamma =\lim_{n\to\infty }\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)$$ I have to show that $$\gamma =\sum_{k=1}^n\frac{1}{k}-\ln(n)-\frac{1}{2n}+\frac{1}{12n^2}-\frac{1}{120n^4}+\cdots$$ but I don't see where come from $$-\frac{1}{2n}+\frac{1}{12n^2}-\frac{1}{120n^4}+\cdots $$ I know that $$\ln(n+1)=-\ln\left(\frac{1}{n}\right)+\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-\frac{1}{4n^4}+\cdots$$ but I can't conclude.

Answer 1

What you have to show in fact boils down to the asymptotic expansion of the digamma function $\psi^{(0)}(x)$ at $x=n+1$, related to the $n$-th Harmonic number: $$\begin{align} H_n=\gamma+\psi^{(0)}(n+1)&\sim\gamma+\log (n+1)-\frac{1}{2(n+1)}+\sum_{k=1}^\infty\frac{\zeta(1-2k)}{(n+1)^{2k}} \\ &\sim \gamma+\log n-\frac{1}{2n}+\sum_{k=1}^\infty\frac{\zeta(1-2k)}{n^{2k}}. \end{align}$$

You can achieve this using the Euler-Maclaurin formula.

Answer 2

This is a consequence of the Euler-Maclaurin sum formula: https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula

This is used to compute the sum of a function if the integral of the function (in this case, $1/x$) and its derivatives are known.

It is, to put it mildly, extremely useful.