Given a/(k+1)! -b/k! +c/(k-1)!=(k^2-k-1)/(k+1)!, what are natural numbers a,b,c?

We'll write the
denominators:

 (k+1)! =
(k-1)!*k*(k+1)

k! =
(k-1)!*k

We'll re-write the identity from
enunciation:

a/(k-1)!*k*(k+1)  - b/(k-1)!*k + c/(k-1)! =
(k^2-k-1)/(k-1)!*k*(k+1)

We'll multiply both sides by
(k-1)!*k*(k+1):

a - b(k+1) + c*k(k+1) =
(k^2-k-1)

We'll remove the
brackets:

a - bk  - b + ck^2 + ck = k^2 - k -
1

We'll combine like terms from the left
side:

ck^2 + k(c - b) + a - b = k^2 - k -
1

Comparing, we'll
get:

c =
1

c - b = -1

b
= c + 1 => b  = 2

a - b
= -1

a = b - 1

a
= 1 

The natural numbers a,b,c
are: a = 1, b  = 2 and c = 1, so
that:

1/(k+1)! 
- 2/k! + 1/(k-1)! =
(k^2-k-1)/(k+1)!