# How would calculate the sum S=`sum_(p=1)^n` (p-1)!/(p+1)!?

Luca B. | Certified Educator

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You need to write the denominator in terms of numerator, using the following factorial relation, such that:

`(p+1)! = (p-1)!*p*(p+1)`

Replacing `(p-1)!*p*(p+1)` for `(p+1)!` in the given summation, yields:

`S = sum_(p=1)^n ((p-1)!)/((p-1)!*p*(p+1))`

Reducing duplicate factors yields:

`S = sum_(p=1)^n 1/(p(p+1))`

You need to use partial fraction expansion, such that:

`1/(p(p+1)) = a/p + b/(p+1)`

`1 = a(p+1) + bp`

`1 = ap + bp + a`

`1 = p(a+b) + a`

Equating the coefficients of like powers, yields:

`{(a + b = 0),(a = 1):} => a = -b => b = -1`

`1/(p(p+1)) = 1/p - 1/(p+1)`

Hence, replacing the difference `1/p - 1/(p+1)` for the fraction `1/(p(p+1))` , yields:

`S = sum_(p=1)^n (1/p - 1/(p+1))`

`S = sum_(p=1)^n (1/p) - sum_(p=1)^n (1/(p+1))`

You need to give values to p from 1 to n, such that:

`p = 1 => 1/1 - 1/2`

`p = 2 => 1/2 - 1/3`

`p = 3 => 1/3 - 1/4`

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

`p = n => 1/n - 1/(n+1)`

Hence, evaluating the summation of terms, yields:

`S = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +... -1/n + 1/n - 1/(n+1)`

Reducing duplicate factors, yields:

`S = 1 - 1/(n + 1) => S = (n + 1 - 1)//(n + 1)`

`S = n/(n + 1)`

Hence, evaluating the given summation, using factorial relations and partial fraction expansion, yields `S = n/(n + 1)` .