Compute : $$L=\lim_{x\to 1^-} \left(\prod_{n=0}^{\infty} \left(\frac {1+x^{n+1}}{1+x^n}\right)^{x^n}\right)$$
My try:
Writing out the first few terms I noticed that the limit can be expressed as $$L=\lim_{x\to 1^-} \frac 12\left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1} -x^n}\right)$$ $$L=\frac 12\lim_{x\to 1^-} \left(\prod_{n=1}^{\infty} (1+x^n)^{x^{n-1}}\right)^{1-x}$$
Converting to exponential form I get $$L=\frac 12 \exp {\left(\lim_{x\to 1^-} (1-x)\sum_{n=1}^{\infty} \left(x^{n-1}\ln(1+x^n)\right)\right)}$$
And got stuck here. I did try to use approximation $\ln x\sim x$ hereon to get final answer as $\frac {\sqrt e}{2}$ but I think it was improper to use approximation and hence I believe that the answer I got is flawed. I also noticed that the form I obtained is quite similar to be used for a Riemann Sum but I didn't get a way to tackle the problem using that way.
Can someone please help me with this problem....
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, $$L=\frac{1}{2}\exp\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m(m+1)}=\frac{1}{2}\exp(2\ln 2-1)=\frac{2}{e}.$$