Let us find the power series expansion of \(\ln(x)\) at \(x=5\) That is, to find the coefficients \(c_n (n \ge 0)\) so that
\begin{equation*}
\ln(x) = c_0 + c_1(x-5) + c_2(x-5)^2 + \cdots = \sum_{n=0}^\infty
c_n(x-5)^n
\end{equation*}
Since the power series expansions that we know are all about \(0\text{,}\) first let us translate the center from \(5\) to \(0\text{.}\) We achieve this by making the substitution \(x+5\text{:}\)
\begin{equation*}
\ln(x+5) = \sum_{n=0}^\infty c_n(x+5 -5)^n = \sum_{n=0}^{\infty}c_nx^n
\end{equation*}
So the task is now to find the power series expansion of \(\ln(x+5)\) about \(x=0\text{.}\) To do that we use a property of \(\ln(x)\) and write
\begin{equation*}
\ln(x+5) = \ln(5(1+x/5)) = \ln(5)+\ln(1+x/5)
\end{equation*}
Next, substituting \(x\) by \(x/5\) in the expansion of \(\ln(1+x)\text{,}\) we get for \(-1 \lt x/5 \le 1\text{,}\) i.e. \(-5 \lt x \le 5\text{:}\)
\begin{align*}
\ln(x+5) & = \ln(5) + \ln(1+x/5) \\
& = \ln(5) + \sum_{n=0}^{\infty} (-1)^{n}
\frac{(x/5)^{n+1}}{n+1}\\
& = \ln(5) + \sum_{n=0} \frac{(-1)^n x^{n+1}}{5^{n+1}(n+1)}
\end{align*}
So substituting \(x\) by \(x-5\) into the above, we get for \(-5
\lt x-5 \le 5\text{,}\) i.e. for \(0 \lt x \le 10\text{:}\)
\begin{align*}
\ln(x) & = \ln(5) + \sum_{n=0}^{\infty} \frac{(-1)^n}{5^{n+1}(n+1)}(x-5)^{n+1} \\
& = \ln(5) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n}(x-5)^n
\end{align*}
That is \(c_0 = \ln(5)\) and \(c_n = (-1)^{n-1}/(n5^n)\) for \(n
\ge 1\text{.}\)