$\lim_{n\to +\infty}\left(\frac{\ln x_n}{\sqrt[n]{e}-1}-n\right)$ where $x_n=\sum_{k=0}^n \frac{1}{k!}$ [closed]

Solution 1:

$$\lim_{n\to\infty}\frac{\ln(x_{n})-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}{\frac{1}{n}\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}$$

$$\lim_{n\to\infty}\frac{\ln(x_{n})-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}}{\frac{1}{n}}=\lim_{n\to\infty}\frac{(\ln(x_{n})-1)+(1-\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}})}{\frac{1}{n}}.$$

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

Solution 2:

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?