Putnam 2004, Problem B-5

Evaluate
\[
\lim_{x \to 1^-} \prod_{n=0}^\infty \left( \frac{1 + x^{n+1}}{1 + x^n} \right)^{x^n}.
\]

Solution:
**First solution:** By taking logarithms, we see that the desired limit is $\exp(L)$, where
$$L = \lim_{x \to 1^-} \sum_{n=0}^\infty x^n \left(\ln(1 + x^{n+1}) - \ln(1 + x^n)\right).$$
Now
$$\sum_{n=0}^N x^n \left(\ln(1 + x^{n+1}) - \ln(1 + x^n)\right)$$
$$= \frac{1}{x} \sum_{n=0}^N x^{n+1} \ln(1 + x^{n+1}) - \sum_{n=0}^N x^n \ln(1 + x^n)$$
$$= x^N \ln(1 + x^{N+1}) - \ln 2 + (1/x - 1) \sum_{n=1}^N x^n \ln(1 + x^n);$$
since $\lim_{N \to \infty} (x^N \ln(1 + x^{N+1})) = 0$ for $0 < x < 1$, we conclude that $L = -\ln 2 + \lim_{x \to 1^-} f(x)$, where
$$f(x) = (1/x - 1) \sum_{n=1}^\infty x^n \ln(1 + x^n)$$
$$= (1/x - 1) \sum_{n=1}^\infty \sum_{m=1}^\infty (-1)^{m+1} x^{n+mn}/m.$$

This final double sum converges absolutely when $0 < x < 1$, since
$$\sum_{n=1}^\infty \sum_{m=1}^\infty x^{n+mn}/m = \sum_{n=1}^\infty x^n (-\ln(1 - x^n))$$
$$< \sum_{n=1}^\infty x^n (-\ln(1 - x)),$$
which converges. (Note that $-\ln(1 - x)$ and $-x^n$ are positive.) Hence we may interchange the summations in $f(x)$ to obtain
$$f(x) = (1/x - 1) \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{m+1} x^{(m+1)n}}{m}$$
$$= (1/x - 1) \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m} \left(\frac{x^m(1 - x)}{1 - x^{m+1}}\right).$$