Is it true that $$\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} (x^n - x^{n+1})\log{(1+x^{n+1})} = 0$$
I'm not entirely sure whether or not this sum tends to 0, since the similar sum $\lim_{x \to 1^{-}} \sum_{n=0}^{\infty} x^n - x^{n+1}$ tends to 1 (in fact the sum is 1 for all $0 < x < 1$).
I've tried to see if the dominated convergence theorem applies so that the limit can be moved inside the sum, but the best dominating values I could come up with were bounding $\log(1+x^{n+1})$ by $\log{(2)}$ and then setting $x = \frac{n}{n+1}$ for the $n$th term. This gives $\log{2}\left(1-\frac{1}{n+1}\right)^n\frac{1}{n+1}$ which is approximately $\frac{\log{2}}{e} \frac{1}{n+1}$ whose sum doesn't converge so I can't use the dominated convergence theorem.
I've also tried obtaining bounds using the fact that $\log{(1+x^{n+1})} < x^{n+1}$ and was only able to show that the limit of the sum is less than $\lim_{x \to 1^{-}} \frac{x}{1+x}$, so less than $\frac{1}{2}$.
\begin{align} |x|<1\implies&\phantom{={}}\sum_{n=0}^{\infty}(x^n-x^{n+1})\log(1+x^{n+1}) \\&=\sum_{n=0}^{\infty}(x^n-x^{n+1})\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}x^{m(n+1)} \\&=\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\sum_{n=0}^{\infty}x^{n(m+1)}(x^m-x^{m+1}) \\&=\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\frac{x^m-x^{m+1}}{1-x^{m+1}} \\&=\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}\frac{x^m}{1+x+\ldots+x^m}. \end{align} Compare to this answer of mine. (The limit is equal to $\color{blue}{2\ln 2-1}$.)