Answer: \(u_n=\ln\left(\frac{1+x^{n+1}}{1+x^n}\right).\)

For \(-1\lt x\lt 1\),

\(\sum_{n=1}^{\infty}u_n=-\ln(1+x).\)

For \(x=1\), the sum is \(0\).

Given

\(u_n=\ln(1+x^{n+1})-\ln(1+x^n),\)

combine the logarithms:

\(u_n=\ln\left(\frac{1+x^{n+1}}{1+x^n}\right).\)

The partial sum is

\(S_N=\sum_{n=1}^{N}\left\{\ln(1+x^{n+1})-\ln(1+x^n)\right\}.\)

This telescopes:

\(S_N=\ln(1+x^{N+1})-\ln(1+x).\)

Equivalently,

\(S_N=\ln\left(\frac{1+x^{N+1}}{1+x}\right).\)

If \(-1\lt x\lt 1\), then \(x^{N+1}\to0\), so

\(\sum_{n=1}^{\infty}u_n=\lim_{N\to\infty}S_N=\ln\left(\frac{1}{1+x}\right)=-\ln(1+x).\)

If \(x=1\), then

\(u_n=\ln2-\ln2=0\)

for every \(n\), so the infinite sum is \(0\).