$\lim_{n\to\infty} \sum_{k=1}^n \frac{k!}{n!}$

Question

I'm presented with the limit $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k!}{n!}$

I've got a hunch that it diverges to infinity but I wasn't able the prove that the sum is superior to a series diverging to infinity.

I would like a hint about how I should start.

Thanks a lot.

Answer 1

It does not diverge to infinity. There are $n$ terms. All but the last are less than $\frac 1n$, so the sum is less than $\frac {n-1}n+1 \lt 2$

Answer 2

Hint for the upper bound: For large $n,$

$$n! + (n-1)! + (n-2)! + \cdots + 1! < n! + (n-1)(n-1)!.$$

Answer 3

For any $n\geq 3$ $$\frac{1}{n!}\sum_{k=1}^{n}k! = 1+\frac{1}{n}+\underbrace{\left(\frac{1}{n(n-1)}+\ldots+\frac{1}{n!}\right)}_{(n-2)\text{ terms bounded by }\frac{1}{n(n-1)}}\leq 1+\frac{1}{n}+\frac{n-2}{n(n-1)}$$ is $1+O\left(\frac{1}{n}\right)$, hence the limit of the LHS as $n\to +\infty$ is simply $\color{red}{\large 1}$.