$\lim_{n\to +\infty}\left(\frac{\ln x_n}{\sqrt[n]{e}-1}-n\right)$ where $x_n=\sum_{k=0}^n \frac{1}{k!}$ What is the value of the limit
$$\lim_{n\to +\infty}\left(\frac{\ln x_n}{\sqrt[n]{e}-1}-n\right) $$ where $x_n=\sum_{k=0}^n \frac{1}{k!}$?
I know only that $\sqrt[n]{e}-1\sim 1/n$, and $x_n\to e$. Thank you in advance.
 A: $$\lim_{n\to\infty}\frac{\ln(x_{n})-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}{\frac{1}{n}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}$$
$$\lim_{n\to\infty}\frac{\ln(x_{n})-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}{\frac{1}{n}}=\lim_{n\to\infty}\frac{(\ln(x_{n})-1)+(1-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}})}{\frac{1}{n}}.$$
Now it suffices to show that $$\lim_{n\to\infty}\frac{(\ln(x_{n})-1)}{\frac{1}{n}}=0$$ and $$\lim_{n\to\infty}\frac{(1-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}})}{\frac{1}{n}}=\frac{-1}{2}.$$
Then as both limit exists and is finite we can add them and say that the sum of these convergent sequence is convergent and does so to $-\frac{1}{2}$.
Now to prove that $$\lim_{n\to\infty}\frac{(\ln(x_{n})-1)}{\frac{1}{n}}=0$$ it is perhaps wise to consider Gary's Hint in the comment.
$$-\frac{1}{en!}-O(\frac{1}{n(en!)^{2}})=\frac{\ln(e-\frac{1}{nn!})-1}{\frac{1}{n}}\leq\frac{(\ln(x_{n})-1)}{\frac{1}{n}} \leq\frac{\ln(e+\frac{1}{nn!})-1}{\frac{1}{n}}\\=\frac{1}{en!}+O(\frac{1}{n(n!)^{2}})$$
So $$\lim_{n\to\infty}\frac{(\ln(x_{n})-1)}{\frac{1}{n}}=0$$
And $$1-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}=\frac{1-\frac{(\frac{1}{n}+\frac{1}{2n^{2}}+O(\frac{1}{n^{3}}))}{\frac{1}{n}}}{\frac{1}{n}}=-\frac{1}{2}-O(\frac{1}{n}).$$
So you have the limit is $\frac{-1}{2}$.
Here $O(\cdot)$ denotes the big O notation
A: Hint: Add and subtract $n\log x_n$ to the limit and split it up
$$L = \lim_{n\to\infty}\log x_n\left(\frac{1}{\sqrt[n]{e}-1}-n\right)+n(\log x_n - 1)$$
which you can show with squeeze theorem simplifies to
$$L = \lim_{n\to\infty}\left(\frac{1}{\sqrt[n]{e}-1}-n\right)$$
Can you complete the limit from here?
