Proof: (related to "Sum of Factorials (I)") 

We provide a proof by induction for all natural numbers $n.$

• Base case: $n=0$
  $$\sum_{k=0}^0 k\cdot k !=0\cdot 0 !=0\cdot 1=1-1=(0+1)!-1.$$
• Induction step $n\to n+1$
  • Assume, $$\sum_{k=0}^n k\cdot k !=(n+1) !-1$$ is correct for an $n\ge 0.$
  • Then $$\begin{align}\sum_{k=0}^{n+1} k\cdot k !&=\sum_{k=0}^{n} k\cdot k ! + (n+1)\cdot (n+1) !\nonumber\\
    &=(n+1)!-1 +(n+1)\cdot (n+1) !\nonumber\\
    &=(n+2)!-1.\nonumber\end{align}$$
• Thus, the above sum of factorials holds for all natural numbers $n\in\mathbb N.$

| | | | created: 2020-06-25 09:13:54 | modified: 2020-06-25 09:14:10 | by: | references: [8311]