Prove that $\frac{2} {(0! + 1! + 2!)} + \frac{3}{( 1! + 2! + 3!)} + · · · + \frac{n} {(n − 2)! + (n − 1)! + n! )}= 1 − \frac{1}{n!}$

Question

My attempts: By observation, we understand that P(1) is not defined. Hence we will prove this statement for $n\geq 2.$

P(2):L.H.S $= 2/0!+1!+2!=1/2$.
R.H.S $= 1-1/2!=1/2$. Therefore, P(2) is true.

Let us assume that P (k) is true.
Therefore, $2 /(0! + 1! + 2!) + 3/( 1! + 2! + 3!) + · · · + k /[(k − 2)! + (k − 1)! + k! ]= 1 − 1/k!$

In order to prove that this statement is true for all n >= 2, we need to prove that P(k+1) is true.

L.H.S $= 2 /(0! + 1! + 2!) + 3/( 1! + 2! + 3!) + · · · + k /[(k − 2)! + (k − 1)! + k! ] + (k + 1)/[(k - 1)! + k! + (k + 1)!]. =1 - 1/k! +(k + 1)/[(k - 1)! + k! + (k + 1)!]$

How do I prove that this is equal to R.H.S. i.e.
$1 - 1/(k+1)!$

Answer 1

$$\begin{align}&\phantom{abv}1-\frac1{k!}+\frac{k+1}{(k-1)!+k!+(k+1)!}\\&=1-\frac {(1+k+k(k+1))-k(k+1)}{k!(1+k+k(k+1))}\\&=1-\frac {1+k}{k!(k^2+2k+1)}\\&=1-\frac1{(k+1)!}\end{align}$$

Answer 2

By telescopic summation:

$$t_k=\frac{k}{(k-2)!+(k-1)!+k!}= \frac{k}{(k-2)![1+k-1+k(k-1)]}= \frac{1}{k~(k-2)!}$$ $$\implies t(k)=\frac{k-1}{k(k-1)~(k-2)!}=\frac{k-1}{k!}= \frac{1}{(k-1)!}-\frac{1}{k!}.$$ $$S=\sum_{k=1}^{n} t_k=1-\frac{1}{n!}.$$