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 '13 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? $\endgroup$ – Anonymous Pi Jun 3 '13 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... $\endgroup$ – Anonymous Pi Jun 3 '13 at 16:50
  • $\begingroup$ please don't use titles only containing MathJax, see here... $\endgroup$ – draks ... Jun 6 '13 at 21:33
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! $\endgroup$ – Anonymous Pi Jun 3 '13 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... $\endgroup$ – Anonymous Pi Jun 3 '13 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 '13 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 '13 at 6:10
  • 1
    $\begingroup$ @B.S. - In Chinese it's called "Splitting Summation" (rough translation) $\endgroup$ – Derek 朕會功夫 Jun 7 '14 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.