Hello everyone!

I've been working on analyzing an algorithm and found that it requires performing $\displaystyle N_1=\sum_{i=1}^n(i+1)!$

Now this can also be expressed as $\displaystyle (n+1)!+n!+(n-1)! + \cdots + 2\ + 1$.

Now what is the complexity of the algorithm? Is it $\displaystyle \mathcal{O}(n+1)!$ ?

But is $\displaystyle N_1 = \mathcal{O}(n!)$ ?

Which raises the following question: Is $\displaystyle n!=\Theta((n+1)!)$ ?

Similarly, if $\displaystyle N_2=\sum_{i=1}^n i(i+1)$ then $\displaystyle N_2$ can be expressed as $\displaystyle n\cdot (n+1) + (n-1)\cdot n +\cdots + 2\cdot 3 + 1\cdot 2$.

Now is $\displaystyle N_2=\mathcal{O}(n^2)$ ?

Thanks!