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. :)
 A: By telescoping
$$\sum_{k=1}^n k\times k!=\sum_{k=1}^n \left((k+1)\times k!- k!\right)=\sum_{k=1}^n  ((k+1)!-k!)=(n+1)!-1$$
A: HINT: $k(k!)=(k+1-1)(k!)=(k+1)!-k!$. Now do the summation and most of the terms will cancel.
A: $$\sum_{k=1}^n k\cdot k!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n \left((k+1) k!- k!\right)=$$
$$=\sum_{k=1}^n  ((k+1)!-k!)=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=$$
$$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-\sum_{k=0}^{n-1} (k+1)!=$$
$$=\sum_{k=1}^{n-1} (k+1)!+(n+1)!-(0+1)!-\sum_{k=1}^{n-1} (k+1)!=$$
$$=(n+1)!-1$$
A: 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
$$
\sum_{k=1}^n k\cdot k! = \int_0^\infty e^{-t}(t^{n+1}-1)dt=(n+1)!-1\,.
$$
A: 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.
A: 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.
