Answer
$\Sigma_{k=0}^n \dfrac{(-1)^{k-1}(x-e)^k}{ke^k}$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_n(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
$f(x)=\ln x$
Differentiate w.r.t x
$f'(x)=\dfrac{1}{x}$
$f''(x)=-\dfrac{1}{x^2}$
$f'''(x)=\dfrac{2}{x^3}$
$f''''(x)=\dfrac{-6}{x^4}$
Now $f(e)=0$, $f'(e)=\dfrac{1}{e}$, $f''(e)=-\dfrac{1}{e^2}$, $f'''(e)=\dfrac{2}{e^3}$, $f''''(e)=-\dfrac{6}{e^4}$
Hence the nth Taylor polynomial for a function is:
$\displaystyle 1+\dfrac{1}{e}(x-e)+ .......+\dfrac{(-1)^{n-1}(x-e)^n}{ne^n}=\Sigma_{k=0}^n \dfrac{(-1)^{k-1}(x-e)^k}{ke^k}$
