# Evaluate $\sum_{k=1}^nk\cdot k!$

I discovered that the summation $\displaystyle\sum_{k=1}^n k\cdot k!$ equals $(n+1)!-1$.

But I want a proof. Could anyone give me one please? Don't worry if it uses very advanced math, I can just check it out on the internet. :)

• Are you familiar with mathematical induction?
– anon
Jun 3, 2013 at 16:34
• @anon Sorry, I'm really not that good in proofs, and I don't know the types of proofs. Could you please help? Jun 3, 2013 at 16:40
• I want to say... THANKS! I finally have enough reputation to vote answers! Well, actually, only to upvote, but I don't think I will be downvoting anytime soon, anyways... Jun 3, 2013 at 16:50
• please don't use titles only containing MathJax, see here... Jun 6, 2013 at 21:33

HINT: $k(k!)=(k+1-1)(k!)=(k+1)!-k!$. Now do the summation and most of the terms will cancel.

• Thanks for editing! And also thanks for the hint! Jun 3, 2013 at 16:38

$$\sum_{k=1}^n k\cdot k!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n \left((k+1) k!- k!\right)=$$ $$=\sum_{k=1}^n ((k+1)!-k!)=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-\sum_{k=0}^{n-1} (k+1)!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-(0+1)!-\sum_{k=1}^{n-1} (k+1)!=$$ $$=(n+1)!-1$$

• Wow. Suppose that math is not always that easy to understand... Jun 3, 2013 at 17:05
• This is very elegant, but for the sake of clear presentation I would suggest that line 3 be just changing the second sum to $\sum_{k=0}^{n-1}(k+1)!$ and the fourth line be $(n+1)!-(0+1)!$ alone. Jun 6, 2013 at 21:32

By telescoping $$\sum_{k=1}^n k\times k!=\sum_{k=1}^n \left((k+1)\times k!- k!\right)=\sum_{k=1}^n ((k+1)!-k!)=(n+1)!-1$$

• I am wonder if Leonardo da Vinci didn't invent telescope, what would this series be called!! :D for my sleepy brother. Dec 5, 2013 at 6:10
• @B.S. - In Chinese it's called "Splitting Summation" (rough translation) Jun 7, 2014 at 5:46

This might be a bit overkill, but I think it's still worth showing: using that $$k!=\int_0^\infty e^{-t}t^kdt\,,$$ we have \begin{aligned} \sum_{k=1}^n k\cdot k!&=\int_0^\infty e^{-t}\sum_{k=1}^n kt^k dt\\ &=\int_0^\infty e^{-t}\frac{d}{dt}\sum_{k=0}^n t^k dt\\ &=\int_0^\infty te^{-t}\frac{d}{dt}\frac{1-t^{n+1}}{1-t}dt \end{aligned} and integrating by parts yields $$\sum_{k=1}^n k\cdot k! = \int_0^\infty e^{-t}(t^{n+1}-1)dt=(n+1)!-1\,.$$

Let's prove by induction that $\sum\limits_{k=1}^{n}{k}\cdot{k!}=(n+1)!-1$.

First, show that this is true for $n=1$:

• $\sum\limits_{k=1}^{1}{k}\cdot{k!}=(1+1)!-1$

Second, assume that this is true for $n$:

• $\sum\limits_{k=1}^{n}{k}\cdot{k!}=(n+1)!-1$

Third, prove that this is true for $n+1$:

• $\sum\limits_{k=1}^{n+1}{k}\cdot{k!}=$

• $\color{red}{\sum\limits_{k=1}^{n}{k}\cdot{k!}}+{(n+1)}\cdot{(n+1)!}=$

• $\color{red}{(n+1)!-1}+{(n+1)}\cdot{(n+1)!}=$

• $(n+1)!\cdot(n+2)-1=$

• $(n+2)!-1$

Please note that the assumption is used only in the part marked red.

Let the $$k^\text{th}$$ term of the given series be $$T_k$$,

$$T_k=k \cdot k!=(k+1-1) \cdot k!=(k+1) \cdot k!-k!=(k+1)!-k!$$

Put $$k=1,T_1=2!-1!$$

Put $$k=2,T_2=3!-2!$$

Put $$k=3,T_3=4!-3!$$

.

.

.

and so on...

Put $$k=n-1, T_{n-1}=n!-(n-1)!$$

Put $$k=n, T_n=(n+1)!-n!$$

Note that $$T_1+T_2+T_3+\dots+T_n=(n+1)!-1$$ since alternate terms one the RHS's cancel out.

Hope this helps.