The Euler-Mascheroni constant as a series

Convergence. $\ln\left(1+\frac1n\right)=\frac1n-\frac1{2n^2}+O\!\left(\frac1{n^3}\right)$, so the general term is $\sim\frac1{2n^2}$: the series converges.

Partial sums. Since $\ln\left(1+\frac1n\right)=\ln(n+1)-\ln n$,
$$\sum_{n=1}^N\left(\frac1n-\ln\left(1+\frac1n\right)\right)=H_N-\ln(N+1).$$

Limit. $H_N-\ln(N+1)=\bigl(H_N-\ln N\bigr)-\ln\left(1+\frac1N\right)\to\gamma-0$, where $\gamma=\lim(H_N-\ln N)\approx0.5772$ is the Euler-Mascheroni constant.

Conclusion. The sum equals $\gamma$.

Remark. This turns the *definition* of $\gamma$ as a limit into a convergent series, and the series converges fast: its terms behave like $\frac1{2n^2}$.