# calculate the summation of k-1/k!.Index variable k starts from 1 and stops to n.

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You need to calculate `sum_(k=1)^n` (k-1)/k!

Write the sum as `sum_(k=1)^n` k/k! - `sum_(k=1)^n` 1/k!

Use the factorial property to write the denominator of fraction k/k! = k/(k-1)!*k.

Reduce by k: k/k! = 1/(k-1)!

`sum_(k=1)^n` (k-1)/k! = `sum_(k=1)^n` 1/(k-1)! - `sum_(k=1)^n` 1/k!

`sum_(k=1)^n` 1/(k-1)! = 1 + 1/1! + 1/2! + .... + 1/(n-1)!

`sum_(k=1)^n` 1/k! = 1/1! + 1/2! + 1/3! + ..... + 1/(n-1)! + 1/n!

`sum_(k=1)^n` 1/(k-1)! - `sum_(k=1)^n` 1/k! = 1 + 1/1! + 1/2! + .... + 1/(n-1)! - 1/1! + 1/2! - 1/3! - ..... - 1/(n-1)! - 1/n!

Reduce opposite terms:

`sum_(k=1)^n` 1/(k-1)! - `sum_(k=1)^n` 1/k! = 1 - 1/n!

**Therefore, the summation gives: `sum_(k=1)^n` (k-1)/k! = 1 - 1/n!**