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

Then I suggest you learn it ASAP. It is one of the most powerful and fundamental tools in a mathematitian's arsenal! — 5xum, Jun 06 '23 at 07:34
Assume we know it for $n-1$ in place of $n$, then $$ \sum\limits_{m = 1}^n {m \cdot m!} = n \cdot n! + \sum\limits_{m = 1}^{n - 1} {m \cdot m!} = \ldots $$ Can you finish? — Gary, Jun 06 '23 at 07:34
Will try, but I have no reference to a good educational material for induction. Found this https://web.stanford.edu/class/archive/cs/cs103/cs103.1142/lectures/03/Small03.pdf — Luthier415Hz, Jun 06 '23 at 07:36
https://www.studysmarter.co.uk/explanations/math/pure-maths/proof-by-induction/ — Gary, Jun 06 '23 at 07:38
@Luthier415Hz Your profile says you are doing a bachelor's in mathematics. If so, the basic calculus textbook you are using, as well as the professor teaching you, are both good sources! — 5xum, Jun 06 '23 at 07:39
I do it from home, I have no books for that here. It is in a foreign country, that is a distance based bachelors. — Luthier415Hz, Jun 06 '23 at 07:41
@Luthier415Hz So you are doing a bachelors in mathematics and you do not have a single book on basic calculus? How does that work? — 5xum, Jun 06 '23 at 07:42
I have books on Calculus, but never seen "induction" in them, I will double check in Adams. — Luthier415Hz, Jun 06 '23 at 07:43
Does this answer your question? Proving $\sum_{k=1}^n kk!=(n+1)!−1$ - found using an Approach0 search. Note there are also quite a few other duplicates here as well, e.g., Using induction to prove that $\sum_{r=1}^n r\cdot r! =(n+1)! -1$, Prove by induction that $\sum_{i=1}^n i!\times i=(n+1)!-1$ for all $n\in \mathbb{N}$, ... — John Omielan, Jun 06 '23 at 09:25
(cont.) Simplify sum of factorials with mathematical induction, Prove $\sum_{j=1}^{n}{j\cdot j!} = (n+ 1)!−1$, Understanding an inductive proof that $ \sum_{i=1}^n i\times i! = (n +1)!-1$, Proving that $\sum\limits_{k = 1}^n k! \cdot k = (n + 1)! - 1$, etc. There's also AoPS threads, e.g., Find the sum, Factorials, A Proof Involving Factorials, etc. — John Omielan, Jun 06 '23 at 09:32

score 1 · Accepted Answer · answered Jun 06 '23 at 07:38

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.

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

1 Answers1