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

Question

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. :)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

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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 $n+1$:

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

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

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

• $(n+1)!\cdot(n+2)-1=$

• $(n+2)!-1$

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

Answer 6

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

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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.