# Closed Form for Factorial Sum

I came across this question in some extracurricular problem sets my professor gave me: what is the closed form notation for the following sum:

$$S_n = 1\cdot1!+2\cdot2!+ ...+n \cdot n!$$

I tried computing some terms, and the only "vague" thing I noticed was that maybe I should be subtracting a term, but I'm really not sure. I went around looking on StackExchange's archives for a closed form of $S_n = 1!+2!+ ...+ n!$ but that didn't help me with my problem much.

Any pointers?

• It's one less than the next factorial. – coffeemath Oct 16 '14 at 17:23
• This has been asked twice recently. – Git Gud Oct 16 '14 at 17:24
• Sorry, Git Gud, I could not find it in the search, I only found threads discussing the sum of n! for n = 1,...,N – Yuro Doosh Oct 16 '14 at 17:25

You're right about subtracting a term; in fact, there's a (clever) strategy called "telescoping sums" and it's particularly useful here, and you won't need induction to show it. You want terms to cancel out so that you're left with the first and last terms only.

If you want to do it yourself, then stop reading here and meditate on this idea: how can you change what's in the summation notation in order to produce a sequence of numbers such that the "middle" terms cancel out?

If you want the solution, here it is:

Let $n=(n+1)-1$, and then substitute this into your summation notation accordingly: $$S=\sum\limits_{i=1}^{n}((n+1)-1)\cdot n!$$ $$S=\sum\limits_{i=1}^{n}[(n+1)\cdot n!-n!]$$ $$S=\sum\limits_{i=1}^{n}((n+1)!-n!)$$

Working out a few terms and the very last, we immediately see: $$S=2!-1!+3!-2!+4!-3!+...+n!-(n-1)!+(n+1)!-n!$$

Which simplifies to:

$$S=(n+1)!-1$$

We can write the above relation as below:

$\sum_{k=1}^{n}k.k!=\sum_{k=1}^{n}(k+1-1)k!=\sum_{k=1}^{n}(k+1)!-\sum_{k=1}^{n}k!=\sum_{k=2}^{n+1}k!-\sum_{k=1}^{n}k!=\sum_{k=1}^{n+1}k!-\sum_{k=1}^{n}k!-1=(n+1)!-1$

Add 1 to $1!$ and you get $2!$, so then carry that over and add that $2 \cdot 2!$ and you get $3!$, so then carry that over and add that to $3 \cdot 3!$ and you get $4!$, and so forth. In the end you will simply get $(n+1)!$. So that's what happens when you add 1.

Let $p$ be a positive integer. We answer a more general question. Is the sum $$S_p(n) = \sum\limits_{k=0}^{n-1} k \cdot (k+p)!$$ given as a hypergeometric term plus a constant. We will be using Gosper's algorithm . Denote $t_n := n \cdot (n+p)!$. Calculate the ratio of the terms in the sum: $$r_k = \left(k+p+1\right) \cdot \frac{k+1}{k}$$ We immediately see that $a_k = k+p+1$, $b_k = 1$ and $c_k = k$ where $$r_k = \frac{a_k}{b_k} \cdot \frac{c_{k+1}}{c_k}$$ and $gcd(a_k,b_{k+h}) = 1$ for all $h=1,2,3,..$. We seek polynomial solutions to the following recurrence: $$(k+p+1) x_{k+1} - 1 \cdot x_k = k$$ Now, if a polynomial solution exist then its degree must be equal to $d = deg(c_k) - max(deg(a_k),deg(b_k))=1-1=0$. Inserting $x_k = A$ we get: $$(A-1) k + p A = 0$$ which gives $A=1$ and $p=0$. Therefore it is only if $p=0$ that the sum is given in closed form and it then reads $S_0(n) = t_n \cdot b_{n-1}/c_n \cdot x_n = n!$