I found that the Taylor series for $h(t)=\ln(t)$ centered at $t_0=1$ is $$\sum^\infty_{n=0}(-1)^n\frac{1}{n+1}(t-1)^{n+1}$$ Now I want to find the interval of confidence for this series, so I use the ratio test: \begin{align} &\lim_{n\rightarrow\infty}\left|\frac{(-1)^{n+1}\frac{1}{n+2}(t-1)^{n+2}}{(-1)^n\frac{1}{n+1}(t-1)^{n+1}}\right|\\ =&|t-1|\lim_{n\rightarrow\infty}\left|(-1)\frac{n+1}{n+2}\right|\\ =&|t-1|\lim_{n\rightarrow\infty}|-1|\\ =&|t-1|=L \end{align} The series converges when $L<1$, so we get the interval $0<t<2$. However, this is incorrect. I was told that the correct interval is $0<t\leq2$.
Why is the interval inclusive of $2$?
The series converges for $|t-1| <1$ and diverges for $|t-1| >1$. For $|t-1|=1$ you have to check separately for convergence. When $t=0$ the series is divergent by comparison with the harmonic series. For $t=2$ it is convergent by the Alternating Series Test.
In general the ratio test does not say anything about the convergence of $\sum a_n (t-c)^{n}$ when $|t-c|=L$ where $L$ is the number you get from the Ratio Test.