Recurrence relation: $c_{k+1}=c_k+\frac{1}{(k+1)!}$

Question

I have no idea how to proceed solving a recurrence relation like this. I know that the terms approach $e$ but beyond that I have no idea. The relation is$$c_{k+1}=c_k+\frac{1}{(k+1)!} \ \ ; \ c_0=1$$

Addendum. I would like an answer in any form other than the sum form.

Answer 1

You may write $$ c_{k+1}-c_k=\frac{1}{(k+1)!} $$ then sum from $k=0$ to $k=n$ to obtain, by telescoping, $$ c_{n+1}-c_0=\sum_{k=0}^n\frac{1}{(k+1)!} $$ or $$ \bbox[20px,border:1px solid #FF9933]{c_n=\sum_{k=0}^n\frac{1}{k!}, \quad n\geq0.} \tag1 $$

If you want an expression without a sum, you may write

$$ \bbox[20px,border:1px solid #FF9933]{c_n=\frac{\left\lfloor n!\:e\right\rfloor}{n!}, \quad n\geq1.} \tag2 $$

where $e$ is the real number such that $\ln e=1$ and where $\left\lfloor x\right\rfloor$ stands for the integer part of $x$.