Answer
$1+\Sigma_{k=1}^{\infty} \dfrac{(-1)^{k-1}(x-e)^k}{ke^k}(x-e)^k$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\ln x$
Differentiate
$f'(x)=\dfrac{1}{x}$
$f''(x)=\dfrac{-1}{x^2}$
$f'''(x)=\dfrac{2}{x^3}$
Now $f(e)=1$ , $f'(e)=\dfrac{1}{e}$ , $f''(e)=\dfrac{-1}{e^2}$, $f'''(e)=\dfrac{2}{e^3}$
Hence the Maclaurin Series for the function is
$\displaystyle 1+\dfrac{1}{e}(x-e)-\dfrac{1}{2e^2}(x-e)^2 .......+ \dfrac{(-1)^{n-1}(x-e)^n}{ne^n} =1+\Sigma_{k=1}^{\infty} \dfrac{(-1)^{k-1}(x-e)^k}{ke^k}(x-e)^k$
