Find the series' radius of convergence. \[ \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{\ln (n+1)} \] A) 1 B) $\infty$, formall $x$ C) 0 D) 2

Step 1 ：The series given is \(\sum_{n=1}^{\infty} \frac{(x-1)^{n}}{\ln (n+1)}\). We want to find the radius of convergence of this series.

Step 2 ：We can use the Ratio Test to find the radius of convergence. The Ratio Test states that if \(\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L\), then the series converges if \(L < 1\), diverges if \(L > 1\), and is inconclusive if \(L = 1\).

Step 3 ：In this case, \(a_n = \frac{(x-1)^{n}}{\ln (n+1)}\). So, \(a_{n+1} = \frac{(x-1)^{n+1}}{\ln (n+2)}\).

Step 4 ：We calculate the limit \(\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{(x-1)^{n+1}/\ln (n+2)}{(x-1)^{n}/\ln (n+1)}\right|\).

Step 5 ：Simplify the above expression to get \(\lim_{n\to\infty} \left|\frac{(x-1)\ln (n+1)}{\ln (n+2)}\right|\).

Step 6 ：As \(n\to\infty\), the limit becomes \(\left| x-1 \right|\).

Step 7 ：For the series to converge, we need \(\left| x-1 \right| < 1\). This implies that \(x\) is in the interval \((0, 2)\).

Step 8 ：Therefore, the radius of convergence is \(\boxed{1}\).