Partial sum using telescopic series

Question

The summation of $(n^2+1)\cdot n!$ using to $N$ using telescopic series.

Question: Find a closed expression in $n$ for: \begin{gather} \text{Let } u_n = (n^2+1)\cdot n! \\ \sum_{n=1}^{N}u_n = \sum_{n=1}^{N}(n^2+1)\cdot n!\\ \end{gather}

My Attempt I tried to split $(n^2+1)\cdot n!$ into two different terms that would telescope.

First try \begin{gather} (n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\ = (n+1)\cdot (n+1)! - 2n\cdot n!\\ \sum_{n=1}^{N}(n+1)\cdot (n+1)! - 2n\cdot n!\\ (2)\cdot(2)! - 2(1)\cdot(1)!\\ (3)\cdot(3)! - 2(2)\cdot(2)! \\ \vdots \\ (n+1)\cdot (n+1)! - 2(n)\cdot (n)!\\ \end{gather} The $2$ on the second term is quite annoying and needless to say it does not work.

Second try \begin{gather} (n^2+1)\cdot n! = ((n+1)^2-2n)\cdot n! \\ = (n+1)\cdot (n+1)! - 2n\cdot n! = (n+1)\cdot(n+1)! -2[(n+1)! - (n)!]\\ \end{gather} Didn't work.

Try #3

\begin{gather} (n+1)\cdot(n+1)! -2[(n+1)! - (n)!] = (n+1)!\cdot(n-1)+2n!\\ \end{gather} This one didn't help at all.

Could anyone please suggest a hint to arrive at the right telescoping expression?

Answer 1

The key is that $$(n^2+1)\cdot n! = n\cdot(n+1)! - (n-1)\cdot n!.$$ Thus $$\sum_{n=1}^Nu_n=(N\cdot(N+1)!-(N-1)\cdot N!) + ((N-1)\cdot N! - (N-2)\cdot (N-1)!) +\\+ \cdots + (1\cdot2!-0\cdot1!) = N\cdot(N+1)!.$$

Answer 2

Hint: let $b_n := n[(n+1)!]$ and compute $b_n - b_{n-1}$.