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

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### 2 Answers

k! = 1*2*3*...*k

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

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

The sum of (k - 1)/k! for k = 1 to n is:

1/0! - 1/1! + 1/1! - 1/2! + 1/2! - 1/3! + ... + 1/(n - 1)! - 1/n!

=> 1/0! - 1/n!

substitute 0! = 1

=> 1/1 - 1/n!

=> (n! - 1)/n!

**The required sum is (n! - 1)/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!.**