Show that $\sum_{m=1}^nm\cdot m!=(n+1)!-1 \forall n\ge 1$

Question

I want to show that

$\sum_{m=1}^nm\cdot m!=(n+1)!-1 \forall n\ge 1$, however I am not sure how to do that.

Let $m!=m(m-1)!$ so we replace it in for $m!$ and get

$\sum_{m=1}^nm\cdot m(m-1)! $

which gives

$\sum_{m=1}^nm^2\cdot(m-1)! $

Then let $(m-1)!=(m-1)(m-2)!$ and we obtain

$\sum_{m=1}^nm^2\cdot(m-1)(m-2)! $

but I am still far from the answer. Any ideas on the next steps?

Thanks

Answer 1

This looks like a typical statement that can be proven by induction.

Base case, $n=1$ is trivial, since both sides of the equation equal $1$.

Induction step:

Assume that the statement is true for some $n$. Then

$$\begin{align} \sum_{m=1}^{n+1} m\cdot m! &= (n+1) (n+1)! + \sum_{m=1}^n m\cdot m! & \text{(by definition of finite sums)}\\ & = (n+1)(n+1)! + (n+1)! - 1&\text{(by the induction hypothesis)}\\ & = (n+1)!\cdot (n+1+1) - 1&\text{(algebraic manipulation)}\\ &= (n+2)\cdot (n+1)! - 1 &\text{(algebraic manipulation)}\\ &= (n+2)! -1&\text{(algebraic manipulation)} \end{align}$$

which proves the statement for $n+1$, concluding the proof.