Let's consider $$U = \sum_{n=1}^{+\infty} (-1)^n\ln(1+\frac{1}{n})$$
Now,
$$U_N = \sum_{n=1}^{2N+1} (-1)^n \ln(\frac{n+1}{n}) = \sum_{k=1}^N \ln \frac{2k+1}{2k} - \sum_{k=0}^N \ln \frac{2k+2}{2k+1}$$
Using $\ln(ab)=\ln a + \ln b$, I obtain two telescoping products and I find
$$U_N = \ln \frac{2N+1}{2} - \ln (2N+2) = \ln \frac{2N+1}{4N+4} \to \ln \frac{1}{2} $$
Then $U = - \ln 2$.
Yet, using $\ln(\frac{a}{b}) = \ln a - \ln b$, and then Stirling formula, my teacher found with a lot of calculus
$$U = \ln \frac{2}{\pi}$$
Is the first method wrong or has the second one failed?