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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. :)

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  • $\begingroup$ 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

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

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  • $\begingroup$ Thanks for editing! And also thanks for the hint! $\endgroup$ Jun 3, 2013 at 16:38
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$$\sum_{k=1}^n k\cdot k!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n \left((k+1) k!- k!\right)=$$ $$=\sum_{k=1}^n ((k+1)!-k!)=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-\sum_{k=0}^{n-1} (k+1)!=$$ $$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-(0+1)!-\sum_{k=1}^{n-1} (k+1)!=$$ $$=(n+1)!-1$$

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    $\begingroup$ Wow. Suppose that math is not always that easy to understand... $\endgroup$ Jun 3, 2013 at 17:05
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    $\begingroup$ 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
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By telescoping $$\sum_{k=1}^n k\times k!=\sum_{k=1}^n \left((k+1)\times k!- k!\right)=\sum_{k=1}^n ((k+1)!-k!)=(n+1)!-1$$

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    $\begingroup$ 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
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    $\begingroup$ @B.S. - In Chinese it's called "Splitting Summation" (rough translation) $\endgroup$ Jun 7, 2014 at 5:46
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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 $$ \sum_{k=1}^n k\cdot k! = \int_0^\infty e^{-t}(t^{n+1}-1)dt=(n+1)!-1\,. $$

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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.

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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.

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