Finding $\lim\limits_{x\to1^-}\Bigl(\prod\limits_{n=0}^{\infty}\Bigl(\frac{1+x^{n+1}}{1+x^n}\Bigr)^{x^n}\Bigr)$
Solution 1:
Expand $\ln(1+x^n)=\displaystyle\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}x^{mn}$ and change the order of summation (which is admissible due to absolute convergence when $|x|<1$). You get $$(1-x)\sum_{n=1}^{\infty}x^{n-1}\ln(1+x^n)=\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\frac{x^m}{1+x+\ldots+x^m}.$$ Here, it is valid to take the limit of the right-hand side termwise (the convergence is uniform). Therefore, $$L=\frac{1}{2}\exp\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m(m+1)}=\frac{1}{2}\exp(2\ln 2-1)=\frac{2}{e}.$$