10
$\begingroup$

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. :)

$\endgroup$
4
  • $\begingroup$ Are you familiar with mathematical induction? $\endgroup$
    – anon
    Jun 3, 2013 at 16:34
  • $\begingroup$ @anon Sorry, I'm really not that good in proofs, and I don't know the types of proofs. Could you please help? $\endgroup$ Jun 3, 2013 at 16:40
  • $\begingroup$ 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... $\endgroup$ Jun 3, 2013 at 16:50
  • $\begingroup$ please don't use titles only containing MathJax, see here... $\endgroup$
    – draks ...
    Jun 6, 2013 at 21:33

6 Answers 6

18
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thanks for editing! And also thanks for the hint! $\endgroup$ Jun 3, 2013 at 16:38
10
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ Wow. Suppose that math is not always that easy to understand... $\endgroup$ Jun 3, 2013 at 17:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – Loki Clock
    Jun 6, 2013 at 21:32
3
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ I am wonder if Leonardo da Vinci didn't invent telescope, what would this series be called!! :D for my sleepy brother. $\endgroup$
    – Mikasa
    Dec 5, 2013 at 6:10
  • 1
    $\begingroup$ @B.S. - In Chinese it's called "Splitting Summation" (rough translation) $\endgroup$ Jun 7, 2014 at 5:46
1
$\begingroup$

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 $$ \sum_{k=1}^n k\cdot k! = \int_0^\infty e^{-t}(t^{n+1}-1)dt=(n+1)!-1\,. $$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.