Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?

Question

Calculate $$\lim_{n \to \infty} (n!e-[n!e])$$

I think that it will be $\infty $ as $\lim_{n \to \infty} (n!e-[n!e])= \infty - \infty = \infty $.

Is it True/false ??

Answer 1

From $\text{e}=\sum\limits_{k=0}^\infty\,\frac{1}{k!}$, we have $$n!\,\text{e}=\sum_{k=0}^n\,\frac{n!}{k!}+\epsilon_n\,,$$ where $$\epsilon_n:=\sum_{k>n}\,\frac{n!}{k!} < \sum_{k>n}\,\frac{1}{(n+1)^{k-n}}=\frac{1}{n}\,.$$ That is, $$\lfloor n!\,\text{e}\rfloor =\sum_{k=0}^n\,\frac{n!}{k!}\text{ and }n!\,\text{e}-\lfloor n!\,\text{e}\rfloor=\epsilon_n\,.$$ Clearly, $\epsilon_n\to0$ as $n\to\infty$.

Answer 2

Clearly $n!e-[n!e]$ is a number between $0$ and $1$, so the limit, if it exists, is also a number between $0$ and $1$. So it is not $\infty$.

Answer 3

Note: $e= \sum\limits_{i=0}^{\infty} \frac{1}{i!}$.

If you multiply by $n!$, then the terms in the sum corresponding to the indices $i=0, 1, \ldots, n$ become integers, so the do not contribute to the fractional part.

So the answer is the same as $\lim\limits_{n\rightarrow \infty} \sum\limits_{i=n+1}^{\infty} \frac{n!}{i!}$.

It is easy to show that the sum $\sum\limits_{i=n+1}^{\infty} \frac{n!}{i!}$ is at most $\frac{1}{n}$. So the limit is 0.