Evaluate the summation (k-1)/k! if k goes from 1 to n.

We can write the general term of summation in this
manner:

(k-1)/k! = 1/(k-1)! -
1/k!

We notice that 1/(k-1)! - 1/k! = [k! -
(k-1)!]/k!*(k-1)!

We'll write k! =
(k-1)!*k

1/(k-1)! - 1/k! = [(k-1)!*k -
(k-1)!]/k!*(k-1)!

We'll factorize by
(k-1)!:

1/(k-1)! - 1/k! = (k-1)!(k -
1)//k!*(k-1)!

We'll simplify and we'll
get:

1/(k-1)! - 1/k! = (k -
1)//k!

Now, we'll replace k by the values 1 to n, one by
one:

k = 1 => 1/(0)! - 1/1! =1 -
1/1!

k = 2 => 1 -
1/1!

...............................

k
= n => 1/(n-1)! - 1/n!

We'll combine the terms of
the sequence, using addition, and we;ll get:

1 - 1/1! + 1 -
1/1! + .... + 1/(n-1)! - 1/n!

We'll eliminate like
terms:

1 - 1/1! + 1 - 1/1! + .... + 1/(n-1)! - 1/n! = 1 -
1/n!

The result of given summation, if k goes
from 1 to n, is Sum(k - 1)//k! = 1 - 1/n!.